blob_id stringlengths 40 40 | repo_name stringlengths 5 127 | path stringlengths 2 523 | length_bytes int64 22 3.06M | score float64 3.5 5.34 | int_score int64 4 5 | text stringlengths 22 3.06M |
|---|---|---|---|---|---|---|
e1ebae30cf58ae268c07ffa4f088c1b5dc3fe644 | gersongroth/maratonadatascience | /Semana 01/01 - Estruturas Sequenciais/10.py | 121 | 3.921875 | 4 | celsius = float(input("Informe a temperatura em celsius: "))
f = celsius * 9 / 5 + 32
print("%.1f graus Farenheit" % f) |
b2612751a0971f2191f1d65bd3e987ce611e9fb9 | DabicD/Studies | /The_basics_of_scripting_(Python)/Exercise2.py | 853 | 3.8125 | 4 | # Exercise description:
#
# "Napisz program drukujący na ekranie kalendarz na zadany miesiąc dowolnego roku
# (użytkownik wprowadza informację postaci: czerwiec 1997–nazwa miesiąca w języku polskim)."
#
##############################################################################################
import loc... |
b5a42001ab9ec5ec13c1c0538824fdcc1d9e4b83 | MarinaSergeeva/Algorithms | /day02_dijkstra.py | 1,301 | 3.78125 | 4 | import heapq
from math import inf
def dijkstra(graph, source):
visited = set([source])
distances = {v: inf for v in graph}
# parents = {v: None for v in graph}
distances[source] = 0
for (v, w) in graph[source]:
distances[v] = w
# parents[v] = source
vertices_heap = [(w, v) for (... |
578a7c30e7e0df3e7e086223575e1a682f4c200e | MarinaSergeeva/Algorithms | /day12_median_maintenance.py | 1,415 | 3.765625 | 4 | import heapq
class MedianMaintainer:
def __init__(self):
self._minheap = [] # for smallest half of fthe array, uses negated numbers
self._maxheap = [] # for largest half of the array
self.median = None # if similar number of elements - use value form maxheap
def insert_element(self, el... |
023911c5beb0dc2cdb8b29f5f4447b6198a85b33 | MarinaSergeeva/Algorithms | /day07_quicksort.py | 757 | 3.921875 | 4 | def partition(array, low, high):
# uses array[low] element for the partition
pivot = array[low]
repl_index = low
for i in range(low + 1, high):
if array[i] < pivot:
repl_index += 1
array[i], array[repl_index] = array[repl_index], array[i]
array[low], array[repl_index]... |
5d185a1960ee3b49934bf30e2e03a48c5ac09db7 | HSabbir/Python-Challange | /day 1.py | 155 | 4.15625 | 4 | ## Print Multiplication table
number = int(input("Enter Your Number: "))
for i in range(10):
print(str(i+1)+' * '+str(number)+' = '+str(number*(i+1))) |
348084b89a6dda4fd185da2863c05cb4cb3b4a3f | Vrittik/LINEAR_REGRESSION_FROM_SCRATCH | /SLR_By_calc.py | 772 | 3.6875 | 4 | import numpy as np
import matplotlib.pyplot as plt
import pandas as pd
from statistics import mean
from ML_library import mlSkies
dataset=pd.read_csv("Salary_Data.csv")
X=dataset.iloc[:,0].values
y=dataset.iloc[:,1].values
X_train,X_test,y_train,y_test = mlSkies.train_test_split(X,y,split_index=0.2)
X_train=np.array... |
89c3903452e5ee6c194159e3a3311fbe2d6ba04d | davidlowryduda/pynt | /pynt/base.py | 4,086 | 4.03125 | 4 | """
base.py
=======
Fundamental components for a simple python number theory library.
License Info
============
(c) David Lowry-Duda 2018 <davidlowryduda@davidlowryduda.com>
This is available under the MIT License. See
<https://opensource.org/licenses/MIT> for a copy of the license,
or see the home github repo
<htt... |
0685f7856c0b7032a146198548e1db5dc3a0bbad | numblr/glaciertools | /test/treehash/algorithm_test.py | 2,242 | 3.53125 | 4 | #!/usr/bin/python
from pprint import pprint
def next_itr(last):
for i in range(1, last + 1):
yield str(i)
def calculate_root(level, itr):
# Base case level
if level == 0:
return next(itr, None)
left = calculate_root(level - 1, itr)
right = calculate_root(level - 1, itr)
retu... |
5f4b7dc66789528b881bc081632e8b54fc6192f3 | bubblegumsoldier/kiwi | /kiwi-user-manager/app/lib/username_validator.py | 333 | 3.546875 | 4 | import re
username_min_string_length = 5
username_max_string_length = 30
username_regex = "^[a-zA-Z0-9_.-]+$"
def validate(username):
if not username_min_string_length <= len(username) <= username_max_string_length:
return False
if not re.match(username_regex, username):
return False
... |
2280d3663399e1dcd1dc76de2ee713c3416c484d | ash-fu/coursera-algo | /Assign2.py | 1,711 | 3.703125 | 4 | count = 0
def mergeSort(alist):
# print("Splitting ",alist)
global count
# count = 0
if len(alist)>1:
mid = len(alist)//2
lefthalf = alist[:mid]
righthalf = alist[mid:]
mergeSort(lefthalf)
mergeSort(righthalf)
# splitSort(lefthalf, righthalf,count)
... |
f3ae407a822f0cd36fdfb490b705e35cd2a275d6 | SaraZ3964/Python | /pybank.py | 420 | 3.625 | 4 | import pandas as pd
file = "budget_data.csv"
data_df = pd.read_csv(file)
data_df.head()
Months = data_df["Date"].count()
Sum = data_df["Profit/Losses"].sum()
Ave = data_df["Profit/Losses"].mean()
Max = data_df["Profit/Losses"].max()
Min = data_df["Profit/Losses"].min()
print("Months:" + str(Months))
print("Total: " ... |
37238eb1a843fb3a7d5e1d36364bf3f0b1bbd7ee | kylehovey/kylehovey.github.io | /spel3o/files/geohash.py | 2,354 | 3.8125 | 4 | import webbrowser
import math
ImTheMap = input("would you like a map site to look up your coordinates? ")
if ImTheMap == "yes":
print('look up your current location on this website')
webbrowser.open("https://csel.cs.colorado.edu/~nishimot/point.html")
else:
print('''okay then, let's continue''')
Lat = input... |
06abfdc014c7ef45be7f8c1ac53007c49983062c | TheAutomationWizard/learnPython | /pythonUdemyCourse/Concepts/General Concepts/slicing.py | 543 | 4.09375 | 4 | list = [1, 2, 34, 4, 56, 7, 8, 99, 2]
def various_slicing_in_python(list_object):
"""
slicing operations in python
:param list_object:
:return:
slice (Start, end , step)
"""
# Using slice method
print(list_object[slice(0, 4, 1)])
# Using index slicing +==> [start : stop+1 : step... |
4bf52a9a6adb7f222b981d2c48c51cd912324faa | TheAutomationWizard/learnPython | /pythonUdemyCourse/Concepts/OOPS/Inheritance/FiguresExample.py | 1,227 | 3.953125 | 4 | class Quadrilateral:
def __init__(self, length, height):
self.length = length
self.height = height
def quad_area(self):
area_ = self.length * self.height
print(f'Area of Quadrilateral with length {self.length} and height : {self.height} is = {area_}')
return area_
clas... |
18d811c538e2db5846bdf7caf92fbca0eaac8f44 | TheAutomationWizard/learnPython | /pythonUdemyCourse/Concepts/OOPS/Inheritance/PythonicInheritance.py | 1,119 | 4.03125 | 4 | class A(object):
def __init__(self, a, *args, **kwargs):
print('I (A) am called from B super()')
print("A", a)
class B(A):
def __init__(self, b, *args, **kwargs):
print('As per inverted flow, i am called from class A1 super()')
super(B, self).__init__(*args, **kwargs)
p... |
cc1812713296f1e020d7a5d426397c2f54622232 | fanying2015/algebra | /algebra/quadratic.py | 407 | 3.78125 | 4 |
def poly(*args):
"""
f(x) = a * x + b * x**2 + c * x**3 + ...
*args = (x, a, b)
"""
if len(args) == 1:
raise Exception("You have only entered a value for x, and no cofficients.")
x = args[0] # x value
coef = args[1:]
results = 0
for power, c in enum... |
38d2e2e55b3ac7a05246a18367a4c82c4bd95cc8 | BOUYAHIA-AB/DeepSetFraudDetection | /split_data.py | 4,321 | 3.5 | 4 | """Build vocabularies of words and tags from datasets"""
from collections import Counter
import json
import os
import csv
import sys
import pandas as pd
def load_dataset(path_csv):
"""Loads dataset into memory from csv file"""
# Open the csv file, need to specify the encoding for python3
use_python3 = sys... |
c742ba20728912aac3293cf456bc83fe88a588cf | Danisaura/phrasalVerbs | /main.py | 6,114 | 3.703125 | 4 | from random import shuffle
# printing the welcome message
print("\n" + "---------------------------------------------------" + "\n" +
"Welcome! this is a script to practice phrasal verbs." + "\n" + "\n" +
"You will be shown sentences with blank spaces inside," + "\n" +
"try to fill them with the corr... |
e6536e8399f1ceccd7eb7d41eddcc302e3dda66b | guv-slime/python-course-examples | /section08_ex04.py | 1,015 | 4.4375 | 4 | # Exercise 4: Expanding on exercise 3, add code to figure out who
# has the most emails in the file. After all the data has been read
# and the dictionary has been created, look through the dictionary using
# a maximum loop (see chapter 5: Maximum and Minimum loops) to find out
# who has the most messages and print how... |
f93dd7a14ff34dae2747f7fa2db22325e9d00972 | guv-slime/python-course-examples | /section08_ex03.py | 690 | 4.125 | 4 | # Exercise 3: Write a program to read through a mail log, build a histogram
# using a dictionary to count how many messages have come from each email
# address, and print the dictionary.
# Enter file name: mbox-short.txt
# {'gopal.ramasammycook@gmail.com': 1, 'louis@media.berkeley.edu': 3,
# 'cwen@iupui.edu': 5, 'antr... |
995c34fb8474004731ba29407120537d9612529f | tacyi/tornado_overview | /chapter01/coroutine_test.py | 787 | 3.953125 | 4 | # 1.什么是协程
# 1.回调过深造成代码很难维护
# 2.栈撕裂造成异常无法向上抛出
# 协程，可被暂停并且切换到其他的协程运行的函数
from tornado.gen import coroutine
# 两种协程的写法，一种装饰器，一种3.6之后的原生的写法，推荐async
# @coroutine
# def yield_test():
# yield 1
# yield 2
# yield 3
#
# yield from yield_test()
#
# return "hello"
async def yield_test():
yield 1
yield... |
cc7a0230928450b5bb71fa5fa6e57429a6e25882 | lmtjalves/CPD | /scripts/gen_random_big_parse_tests.py | 1,160 | 3.609375 | 4 | #!/bin/python
import sys, argparse, random
def test_rand(t):
if t == "both":
return random.randint(0,1)
elif t == "positive":
return 0
else:
return 1
parser = argparse.ArgumentParser(description="problem gen. clauses might be duplicate and have repeated variables")
parser.add_arg... |
6214901ec8317a2ead9409991548282f5ce33c57 | bobgautier/rjgtoys-config | /examples/translate.py | 590 | 3.890625 | 4 | """
examples/translate.py: translate words using a dictionary
"""
import argparse
import os
from typing import Dict
from rjgtoys.config import Config, getConfig
class TranslateConfig(Config):
words: Dict[str, str]
cfg = getConfig(TranslateConfig)
def main(argv=None):
p = argparse.ArgumentParser()
cf... |
2d7724e5786f00b9f2c1e2f8640ebde7138f7c85 | maryamkh/MyPractices | /Find_Nearest_Smaller_Element.py | 1,930 | 3.90625 | 4 | '''
Given an array, find the nearest smaller element G[i] for every element A[i] in the array such that the element has an index smaller than i.
Elements for which no smaller element exist, consider next smaller element as -1.
Output: An array of prev .smaller value of each item or -1(if no smaller value ex... |
ff2ac738c1718fa12bd84e447ddc9b0e1080420a | maryamkh/MyPractices | /Min_Sum_Path_Bottom_Up.py | 4,255 | 4.375 | 4 | #!usr/bin/python
'''
The approach is to calculate the minimum cost for each cell to figure out the min cost path to the target cell.
Assumpthion: The movementns can be done only to the right and down.
In the recursive approach there is a lot of redundent implementation of the sub-problems. With Dynamic prog... |
bd745cc83163bd91f36ab2f2d034f7f0a02093c0 | maryamkh/MyPractices | /Pow_function_recursive.py | 697 | 3.921875 | 4 | '''
Implement pow(x, n), which calculates x raised to the power n (i.e., xn).
Example:
Input: x = 2.00000, n = -2
Output: 0.25000
Explanation: 2-2 = 1/22 = 1/4 = 0.25
Time Complexity: O(log(n))===> In this solution n is reduce to half and therefore this is the time cimplexity
Spcae Complexity: We need to do the computa... |
d1917f2bd44d327224cfb121c1d18f68c5de2383 | maryamkh/MyPractices | /Stairs.py | 2,092 | 4.09375 | 4 | #!usr/bin/python
'''
Find the numbers of ways we can reach the top of the stairs if we can only clime 1 or 2 steps each time?
Input: Integer: The stair number in which we should reach to
Output: Integer: The numebr of ways we can clime up to reach target stair
Reaching any stair with only 1 step climing... |
f9a66f5b0e776d063d812e7a7185ff6ff3c5615f | maryamkh/MyPractices | /ReverseLinkedList.py | 2,666 | 4.3125 | 4 | '''
Reverse back a linked list
Input: A linked list
Output: Reversed linked list
In fact each node pointing to its fron node should point to it back node ===> Since we only have one direction accessibility to a link list members to reverse it I have to travers the whole list, keep the data of the nodes and then rearra... |
145413092625adbe30b158c21e5d27e2ffcfab50 | maryamkh/MyPractices | /Squere_Root.py | 1,838 | 4.1875 | 4 | #!/usr/bin/python
'''
Find the squere root of a number. Return floor(sqr(number)) if the numebr does not have a compelete squere root
Example: input = 11 ===========> output = 3
Function sqrtBinarySearch(self, A): has time complexity O(n), n: given input: When the number is too big it becomes combursome
... |
55ab2d3473fb7ff9485f1ff835dd599d427e0a5d | hokiespider/win_probability | /win_probability.py | 4,054 | 3.734375 | 4 | #!/usr/bin/env python
# coding: utf-8
import pandas as pd
import requests
import json
# What school are you analyzing?
school = "Virginia Tech"
# Get data from the API
df = pd.DataFrame()
for x in range(2013, 2020, 1): # Line data is only available 2013+
parameters = {
"team": school,
"year": ... |
65256d3f3d66a41bd69be4dc55bb89b2c643036e | DaniRyland-Lawson/CP1404-cp1404practicals- | /prac_06/demo_program.py | 1,784 | 3.75 | 4 | """CP1404 Programming II demo program week 6 prac
0. Pattern based programming
1. Names based on problem domain
2. Functions at the same leve of abstraction( main should "look" the same
Menu- driven program
load products
- L_ist products
- S_wap sale status (get product number with error checking)
- Q_uit (save file)... |
e47f166763aad48f70da971a79953db8875531b7 | DaniRyland-Lawson/CP1404-cp1404practicals- | /prac_07/miles_to_kms.py | 1,154 | 3.53125 | 4 | """CP1404 Programming II Week 7 Kivy - Gui Program to convert Miles to Kilometres."""
from kivy.app import App
from kivy.lang import Builder
from kivy.app import StringProperty
MILES_TO_KM = 1.60934
class MilesToKilometres(App):
output_km = StringProperty()
def build(self):
self.title = "Convert Mi... |
4a20f0a4d156b03c5e658e0073f8086ab5ca0b95 | DaniRyland-Lawson/CP1404-cp1404practicals- | /prac_08/unreliable_car_test.py | 676 | 3.640625 | 4 | """CP1404 Programming II
Test to see of UnreliableCar class works."""
from prac_08.unreliable_car import UnreliableCar
def main():
"""Test for UnreliableCars."""
# Create some cars for reliability
good_car = UnreliableCar("Good Car", 100, 90)
bad_car = UnreliableCar("Bad Car", 100, 10)
# Attemp... |
de7b30a4f51727085b556dc01763180a3fdedffd | Monkin6/yarygin | /hm4.py | 132 | 3.5 | 4 | n = int(input())
maximum = -1
while n != 0:
if n % 10 > maximum:
maximum = n % 10
n = n // 10
print(maximum)
|
c0c510cbebedb03947bb2b9ed16c16efa23a4956 | Sahil-k1509/Python_and_the_Web | /Scripts/Miscellaneous/Email_extractor/extract_emails.py | 407 | 3.78125 | 4 | #!/usr/bin/env python3
import re
print("Enter the name of the input file: ")
file=str(input())
try:
f = open(file,"r")
except FileNotFoundError:
print("File does not exists")
email={}
for i in f:
em = re.findall('\S+@\S+\.\S+',i)
for j in em:
email[j]=email.get(j,0)+1
f.close()
for... |
d99e28fea0f6d213659694f220451d12930dcd84 | Mertvbli/JustTry | /CW_filter_list_7kyu.py | 363 | 3.640625 | 4 | def filter_list(l):
new_list = []
for number in l:
if str(number).isdigit() and str(number) != number:
new_list.append(number)
return new_list # or [number for number in l if isinstance(number, int)
print(filter_list([1,2,'a','b']))
print(filter_list([1,'a','b',0,15]))
print... |
ddbfeec96361f4c3576874c2ff007d88717f1566 | CodingDojoDallas/python_sep_2018 | /austin_parham/product.py | 948 | 3.671875 | 4 | class Product:
def __init__(self,price,item_name,weight,brand):
self.price = price
self.item_name = item_name
self.weight = weight
self.brand = brand
self.status = "for sale"
self.display_info()
def sell(self):
self.status = "sold"
return self
def add_tax(self,x):
self.price = (self.price * x) + ... |
13dac1bd992f843d432a94f266a283671e39c2fa | CodingDojoDallas/python_sep_2018 | /Solon_Burleson/Basics.py | 370 | 3.796875 | 4 | # def allOdds():
# for x in range(3001):
# if x % 2 != 0:
# print (x)
# allOdds()
# def Iterate(arr):
# for x in arr:
# print (x)
# Iterate([1,2,3,4,5])
# def Sumlist(arr):
# sum = 0
# for x in arr:
# sum += x
# return sum
# print(Sumlist([1,2,3,4,5]))
list =... |
bb488183c87ed750f3cd459bda9d758416b5613e | CodingDojoDallas/python_sep_2018 | /austin_parham/func_intermediate_1.py | 819 | 3.828125 | 4 | def randInt():
import random
hold = (random.random()*100)
hold = int(hold)
print(hold)
randInt()
def randInt():
import random
hold = (random.random()*50)
hold = int(hold)
print(hold)
randInt()
def randInt():
import random
hold = (random.uniform(50,100))
hold = int(hold)
print(hold)
randInt()
def randInt(... |
8b9f850c53a2a020b1deea52e301de0d2b6c47c3 | CodingDojoDallas/python_sep_2018 | /austin_parham/user.py | 932 | 4.15625 | 4 | class Bike:
def __init__(self, price, max_speed, miles):
self.price = price
self.max_speed = max_speed
self.miles = miles
def displayInfo(self):
print(self.price)
print(self.max_speed)
print(self.miles)
print('*' * 80)
def ride(self):
print("Riding...")
print("......")
print("......")
self.mi... |
60abefff5fa43ad4a30bee1e102f3a31a08c15b6 | CodingDojoDallas/python_sep_2018 | /albert_garcia/python_oop/slist.py | 1,276 | 3.78125 | 4 | class Node:
def __init__(self, value):
self.value = value
self.next = None
class SList:
def __init__(self, value):
node = Node(value)
self.head = node
def Addnode(self, value):
node = Node(value)
runner = self.head
while (runner.next != None):
... |
189bd9eb0029b856f348e9ff86f32ceb6f99d84b | CodingDojoDallas/python_sep_2018 | /Solon_Burleson/RunCode.py | 380 | 3.703125 | 4 | class MathDojo:
def __init__(self):
self.value = 0
def add(self, *nums):
for i in nums:
self.value += i
return self
def subtract(self, *nums):
for i in nums:
self.value -= i
return self
def result(self):
print(self.value)
x = M... |
ab847b8b4d3b115f88b96b560b41f076a7bd6bdc | CodingDojoDallas/python_sep_2018 | /Solon_Burleson/FunctionsIntermediateI.py | 195 | 3.65625 | 4 | import random
def randInt(max=0, min=0):
if max == 0 and min == 0:
print(int(random.random()*100))
else:
print(int(random.random()*(max-min)+min))
randInt(max=500,min=50)
|
36a4f28b97be8be2e7f6e20965bd21f554270704 | krismosk/python-debugging | /area_of_rectangle.py | 1,304 | 4.6875 | 5 | #! /usr/bin/env python3
"A script for calculating the area of a rectangle."
import sys
def area_of_rectangle(height, width = None):
"""
Returns the area of a rectangle.
Parameters
----------
height : int or float
The height of the rectangle.
width : int or float
The width o... |
dda3e4ff366d47cea012f9bfede9819fac448af9 | BlueAlien99/minimax-reversi | /app/gui/utils.py | 8,441 | 3.515625 | 4 | import enum
import pygame
from typing import List
import pygame.freetype
from pygame_gui.elements.ui_drop_down_menu import UIDropDownMenu
from pygame_gui.elements.ui_text_entry_line import UITextEntryLine
import time
class Color(enum.Enum):
NO_COLOR = -1
BLACK = "#212121"
WHITE = "#f5f5f5"
GREEN = "#3... |
dacaf7998b9ca3a71b6b90690ba952fb56349ab9 | Kanthus123/Python | /Design Patterns/Creational/Abstract Factory/doorfactoryAbs.py | 2,091 | 4.1875 | 4 | #A factory of factories; a factory that groups the individual but related/dependent factories together without specifying their concrete classes.
#Extending our door example from Simple Factory.
#Based on your needs you might get a wooden door from a wooden door shop,
#iron door from an iron shop or a PVC door from th... |
ab049070f8348f4af8caeb601aee062cc7a76af2 | Kanthus123/Python | /Design Patterns/Structural/Decorator/VendaDeCafe.py | 1,922 | 4.46875 | 4 | #Decorator pattern lets you dynamically change the behavior of an object at run time by wrapping them in an object of a decorator class.
#Imagine you run a car service shop offering multiple services.
#Now how do you calculate the bill to be charged?
#You pick one service and dynamically keep adding to it the prices f... |
4bcdaa732a2a499c3e52a902911b1a6cbc6636bf | Kanthus123/Python | /Design Patterns/Behavioral/Strategy/main.py | 817 | 3.671875 | 4 | #Strategy pattern allows you to switch the algorithm or strategy based upon the situation.
#Consider the example of sorting, we implemented bubble sort but the data started to grow and bubble sort started getting very slow.
#In order to tackle this we implemented Quick sort. But now although the quick sort algorithm w... |
478e6714f68fb421aff714cf178486c60d46980b | russellgao/algorithm | /dailyQuestion/2020/2020-05/05-14/python/solution.py | 376 | 3.78125 | 4 | from functools import reduce
# 位运算 写法1
def singleNumber1(nums: [int]) -> int:
return reduce(lambda x, y: x ^ y, nums)
# 位运算 写法2
def singleNumber2(nums: [int]) -> int :
result = nums[0]
for i in range(1,len(nums)) :
result ^= nums[i]
return result
if __name__ == "__main__" :
nums = [2,3,4... |
ba0c5f0469a2b8ef74c669af85355c81c4a40eb6 | russellgao/algorithm | /dailyQuestion/2020/2020-10/10-10/python/solution.py | 891 | 4 | 4 | # Definition for singly-linked list.
class ListNode:
def __init__(self, x):
self.val = x
self.next = None
def partition(head: ListNode, x: int) -> ListNode:
first = first_head = ListNode(0)
second = second_head = ListNode(0)
while head:
if head.val < x:
first.next ... |
a6f65dc2d6ac9f5f160229c2dc76b2d74c150550 | russellgao/algorithm | /dailyQuestion/2020/2020-07/07-29/python/solution_n.py | 263 | 3.890625 | 4 |
# 一次遍历
def missingNumber(nums: [int]) -> int:
for i,v in enumerate(nums) :
if i != v :
return i
return nums[-1] + 1
if __name__ == "__main__" :
nums = [0,1,2,3,4,5,6,7,9]
result = missingNumber(nums)
print(result) |
1587894d5e65ee725de94d02e15cd0ec84f1987b | russellgao/algorithm | /dailyQuestion/2020/2020-06/06-02/python/solution.py | 445 | 3.515625 | 4 |
def lengthOfLongestSubstring(s: str) -> int:
tmp = set()
result = 0
j = 0
n = len(s)
for i in range(n) :
if i != 0 :
tmp.remove(s[i-1])
while j < n and s[j] not in tmp :
tmp.add(s[j])
j += 1
result = max(result, j - i)
return result
i... |
47ddbc6c436d80ff4ba68199da5e803edabf3402 | russellgao/algorithm | /dailyQuestion/2020/2020-05/05-22/python/solution_dlinknode.py | 2,137 | 3.84375 | 4 | # 双向链表求解
class DlinkedNode():
def __init__(self):
self.key = 0
self.value = 0
self.next = None
self.prev = None
class LRUCache():
def __init__(self, capacity: int):
self.capacity = capacity
self.size = 0
self.cache = {}
self.head = DlinkedNode()... |
98786826bd97d037c30c7f2b4244b7101ccd963e | russellgao/algorithm | /data_structure/heap/python/002.py | 1,317 | 4.03125 | 4 | # 堆排序
def buildMaxHeap(lists):
"""
构造最大堆
:param lists:
:return:
"""
llen = len(lists)
for i in range(llen >> 1, -1, -1):
heapify(lists, i, llen)
def heapify(lists, i, llen):
"""
堆化
:param lists:
:param i:
:return:
"""
largest = i
left = 2 * i + 1
... |
77ec5582550e18cce771f24058e78bc18686ec9a | russellgao/algorithm | /dailyQuestion/2020/2020-04/04-29/python/solution.py | 1,683 | 3.84375 | 4 | class ListNode:
def __init__(self, x):
self.val = x
self.next = None
# 方法一
# 递归，原问题可以拆分成自问题，并且自问题和原问题的问题域完全一样
# 本题以前k个listnode 为原子进行递归
def reverseKGroup1(head: ListNode, k: int) -> ListNode:
cur = head
count = 0
while cur and count!= k:
cur = cur.next
count += 1
if c... |
30cea365bbb1ea986b435692edfb5eb4118249cc | russellgao/algorithm | /dailyQuestion/2020/2020-08/08-06/python/solution_dict.py | 1,341 | 3.71875 | 4 | def palindromePairs(words: [str]) -> [[int]]:
indices = {}
result = []
n = len(words)
def reverse(word):
_w = list(word)
n = len(word)
for i in range(n >> 1):
_w[i], _w[n - 1 - i] = _w[n - i - 1], _w[i]
return "".join(_w)
def isPalindromes(word: str, lef... |
32c5ca8e7beb18feafd101e6e63da060c3c47647 | russellgao/algorithm | /data_structure/binaryTree/preorder/preoder_traversal_items.py | 695 | 4.15625 | 4 |
# 二叉树的中序遍历
class TreeNode:
def __init__(self, x):
self.val = x
self.left = None
self.right = None
# 迭代
def preorderTraversal(root: TreeNode) ->[int]:
result = []
if not root:
return result
queue = [root]
while queue:
root = queue.pop()
if root:
... |
5723367d25964f32d4f5bc67a99e3f824309f639 | russellgao/algorithm | /dailyQuestion/2020/2020-10/10-01/python/solution.py | 927 | 4.03125 | 4 | # Definition for a binary tree node.
class TreeNode:
def __init__(self, x):
self.val = x
self.left = None
self.right = None
def increasingBST(root: TreeNode) -> TreeNode:
result = node = TreeNode(0)
queue = []
while root or len(queue) > 0 :
while root :
queue... |
b8189f9da4e8491b8871a75225d4376c9ea2cc0c | russellgao/algorithm | /dailyQuestion/2020/2020-06/06-06/python/solution.py | 476 | 3.796875 | 4 |
def longestConsecutive(nums: [int]) -> int:
nums = set(nums)
longest = 0
for num in nums:
if num - 1 not in nums:
current = num
current_len = 1
while current + 1 in nums:
current += 1
current_len += 1
longest = max(long... |
cb8bdd7d8b00d9b6214787b82fe5766228741eee | russellgao/algorithm | /data_structure/sort/tim_sort.py | 2,031 | 3.734375 | 4 | import time
def binary_search(the_array, item, start, end): # 二分法插入排序
if start == end:
if the_array[start] > item:
return start
else:
return start + 1
if start > end:
return start
mid = round((start + end) / 2)
if the_array[mid] < item:
return... |
a4d555397fb194beb604dd993a6b08c746409046 | russellgao/algorithm | /dailyQuestion/2020/2020-07/07-11/python/solution.py | 546 | 3.828125 | 4 | def subSort(array: [int]) -> [int]:
n = len(array)
first,last = -1,-1
if n == 0 :
return [first,last]
min_a = float("inf")
max_a = float("-inf")
for i in range(n) :
if array[i] >= max_a :
max_a = array[i]
else :
last = i
if array[n-1-i] <= ... |
c836d3a26bb6d7432f734c7a771df38e8aaec095 | russellgao/algorithm | /dailyQuestion/2020/2020-07/07-18/python/solution_recurse.py | 708 | 3.875 | 4 | def isInterleave(s1, s2, s3):
"""
:type s1: str
:type s2: str
:type s3: str
:rtype: bool
"""
m = len(s1)
n = len(s2)
t = len(s3)
if m + n != t:
return False
dp = [[False] * (n + 1) for _ in range(m + 1)]
dp[0][0] = True
for i in range(m + 1):
for j in ... |
bd9249d9d6593d652adca04e69220e3326615cd4 | russellgao/algorithm | /dailyQuestion/2020/2020-06/06-14/python/solution_vertical.py | 432 | 3.96875 | 4 | def longestCommonPrefix(strs: [str]) -> str:
if not strs:
return ""
length, count = len(strs[0]), len(strs)
for i in range(length):
c = strs[0][i]
if any(i == len(strs[j]) or strs[j][i] != c for j in range(1, count)):
return strs[0][:i]
return strs[0]
if __name__ ... |
ff3ce63ef2e076344a7e1226b9fadf5a37a5653b | russellgao/algorithm | /dailyQuestion/2021/2021-03/03-20/python/solution.py | 961 | 3.515625 | 4 | class Solution:
def evalRPN(self, tokens: [str]) -> int:
stack = []
for i in range(len(tokens)) :
tmp = tokens[i]
if tmp == "+" :
num1 = stack.pop()
num2 = stack.pop()
stack.append(num2 + num1)
elif tmp == "-" :
... |
9d4fa8524fe6b0b3172debd49bf081e98f5a0282 | russellgao/algorithm | /dailyQuestion/2020/2020-06/06-09/python/solution_mod.py | 335 | 3.6875 | 4 | # 动态求余法
def translateNum(num: int) -> int:
f_1 = f_2 = 1
while num:
pre = num % 100
f = f_1 + f_2 if pre >= 10 and pre <= 25 else f_1
f_2 = f_1
f_1 = f
num = num // 10
return f_1
if __name__ == '__main__' :
num = 12258
result = translateNum(num)
print(res... |
861fab844f5dcbf86c67738354803e27a0a303e9 | russellgao/algorithm | /dailyQuestion/2020/2020-05/05-31/python/solution_recursion.py | 950 | 4.21875 | 4 | # Definition for a binary tree node.
class TreeNode:
def __init__(self, x):
self.val = x
self.left = None
self.right = None
# 递归
def isSymmetric(root: TreeNode) -> bool:
def check(left, right):
if not left and not right:
return True
if not left or not right:
... |
21f1cf35cd7b3abe9d67607712b62bfa4732e4ce | russellgao/algorithm | /dailyQuestion/2020/2020-05/05-01/python/solution.py | 944 | 4.125 | 4 | # Definition for singly-linked list.
class ListNode(object):
def __init__(self, x):
self.val = x
self.next = None
def reverseList(head):
"""
递归 反转 链表
:type head: ListNode
:rtype: ListNode
"""
if not head :
return None
if not head.next :
return head
l... |
86bee5da92c7b029a182b67d9e3aa4bb2eb1b949 | franztrierweiler/maths_python | /exercices_02mars.py | 2,000 | 3.8125 | 4 | from fractions import *
def somme():
S = 1
for i in range (1,21):
S = S + pow(Fraction(1,3),i)
# Converts fraction to float
# and returns the result
return float(S);
def compte_espaces(phrase):
nbre = 0
for caractere in phrase:
if caractere==" ":
nbre=n... |
407273956e8e87a116912ee44dc192e8233f5fac | swainsubrat/Haw | /Dependencies/DataFrameBuilder.py | 5,313 | 3.9375 | 4 | """
Structures dataframes for plotting
"""
import re
import pandas as pd
from io import StringIO
from pandas.core.frame import DataFrame
def basicDataFrameBuilder(FILE: StringIO) -> DataFrame:
"""
Function to pre-process the raw text file and format it
to get a dataframe out of it
1. Datetime extr... |
fc94459d32944d0e67d1870d0b2e864263dc8319 | Narusi/Python-Kurss | /Uzdevums Lists.py | 3,209 | 4.1875 | 4 | #!/usr/bin/env python
# coding: utf-8
# # Klases Uzdevumi - Lists
# ## 1.a Vidējā vērtība
# Uzrakstīt programmu, kas liek lietotājam ievadīt skaitļus(float).
# Programma pēc katra ievada rāda visu ievadīto skaitļu vidējo vērtību.
# PS. 1a var iztikt bez lists
#
# 1.b Programma rāda gan skaitļu vidējo vērtību, gan V... |
d278d8c5efedbd61317118887461b052690dd605 | wulinlw/leetcode_cn | /常用排序算法/quicksort.py | 828 | 3.8125 | 4 | #!/usr/bin/python
#coding:utf-8
# 快速排序
def partition(arr,low,high):
i = ( low-1 ) # 最小元素索引
pivot = arr[high]
for j in range(low , high):
# 当前元素小于或等于 pivot
if arr[j] <= pivot:
i = i+1
arr[i],arr[j] = arr[j],arr[i]
# print(i,arr)
arr[i+1],arr[... |
8beaa095846c553f6c970e062494b068733a5d6a | wulinlw/leetcode_cn | /leetcode-vscode/671.二叉树中第二小的节点.py | 2,205 | 3.734375 | 4 | #
# @lc app=leetcode.cn id=671 lang=python3
#
# [671] 二叉树中第二小的节点
#
# https://leetcode-cn.com/problems/second-minimum-node-in-a-binary-tree/description/
#
# algorithms
# Easy (45.43%)
# Likes: 61
# Dislikes: 0
# Total Accepted: 8.5K
# Total Submissions: 18.6K
# Testcase Example: '[2,2,5,null,null,5,7]'
#
# 给定一个非空... |
8504344ee52ab7c26d4e0a926e97ad8a8874308f | wulinlw/leetcode_cn | /程序员面试金典/面试题01.05.一次编辑.py | 1,707 | 3.75 | 4 | #!/usr/bin/python
#coding:utf-8
# 面试题 01.05. 一次编辑
# 字符串有三种编辑操作:插入一个字符、删除一个字符或者替换一个字符。 给定两个字符串，编写一个函数判定它们是否只需要一次(或者零次)编辑。
# 示例 1:
# 输入:
# first = "pale"
# second = "ple"
# 输出: True
# 示例 2:
# 输入:
# first = "pales"
# second = "pal"
# 输出: False
# https://leetcode-cn.com/problems/one-away-lcci/
from typing import Lis... |
539c0e8e5f78fb080ec39bf80e69aa14161cbc3c | wulinlw/leetcode_cn | /剑指offer/55_2_平衡二叉树.py | 1,298 | 3.609375 | 4 | #!/usr/bin/python
#coding:utf-8
# // 面试题55（二）：平衡二叉树
# // 题目：输入一棵二叉树的根结点，判断该树是不是平衡二叉树。如果某二叉树中
# // 任意结点的左右子树的深度相差不超过1，那么它就是一棵平衡二叉树。
class TreeNode:
def __init__(self, x):
self.val = x
self.left = None
self.right = None
class Solution:
# 归并的套路，先拿到子结果的集，在处理资结果的集
def IsBalanced(self, r... |
28cb62abf374b9a10b964d944b5b858474c2422c | wulinlw/leetcode_cn | /leetcode-vscode/872.叶子相似的树.py | 1,729 | 3.921875 | 4 | #
# @lc app=leetcode.cn id=872 lang=python3
#
# [872] 叶子相似的树
#
# https://leetcode-cn.com/problems/leaf-similar-trees/description/
#
# algorithms
# Easy (62.23%)
# Likes: 49
# Dislikes: 0
# Total Accepted: 9.7K
# Total Submissions: 15.5K
# Testcase Example: '[3,5,1,6,2,9,8,null,null,7,4]\n' +
# '[3,5,1,6,7,4,2,n... |
efcf849ffdf2209df24b021a2dff59c5138618aa | wulinlw/leetcode_cn | /程序员面试金典/面试题05.04.下一个数.py | 5,329 | 3.515625 | 4 | # #!/usr/bin/python
# #coding:utf-8
#
# 面试题05.04.下一个数
#
# https://leetcode-cn.com/problems/closed-number-lcci/
#
# 下一个数。给定一个正整数，找出与其二进制表达式中1的个数相同且大小最接近的那两个数（一个略大，一个略小）。
# 示例1:
#
#
# 输入：num = 2（或者0b10）
# 输出：[4, 1] 或者（[0b100, 0b1]）
#
#
# 示例2:
#
#
# 输入：num = 1
# 输出：[2, -1]
#
#
# 提示:
#
#
# num的范围在[1, 21... |
c8eab7467ee25294d227d9d16ef6cea5f97d7ab2 | wulinlw/leetcode_cn | /程序员面试金典/面试题02.07.链表相交.py | 3,147 | 3.6875 | 4 | #!/usr/bin/python
#coding:utf-8
# 面试题 02.07. 链表相交
# 给定两个（单向）链表，判定它们是否相交并返回交点。请注意相交的定义基于节点的引用，而不是基于节点的值。换句话说，如果一个链表的第k个节点与另一个链表的第j个节点是同一节点（引用完全相同），则这两个链表相交。
# 示例 1：
# 输入：intersectVal = 8, listA = [4,1,8,4,5], listB = [5,0,1,8,4,5], skipA = 2, skipB = 3
# 输出：Reference of the node with value = 8
# 输入解释：相交节点的值为 8 （注意，如果... |
b282517a21d331b04566b1697602e99244800f48 | wulinlw/leetcode_cn | /leetcode-vscode/15.三数之和.py | 2,939 | 3.546875 | 4 | #
# @lc app=leetcode.cn id=15 lang=python3
#
# [15] 三数之和
#
# https://leetcode-cn.com/problems/3sum/description/
#
# algorithms
# Medium (25.70%)
# Likes: 2228
# Dislikes: 0
# Total Accepted: 241.9K
# Total Submissions: 874.9K
# Testcase Example: '[-1,0,1,2,-1,-4]'
#
# 给你一个包含 n 个整数的数组 nums，判断 nums 中是否存在三个元素 a，b，c... |
c1de2f72e8609e27c4c06ec7c843559d6ae6e447 | wulinlw/leetcode_cn | /程序员面试金典/面试题08.03.魔术索引.py | 1,651 | 3.84375 | 4 | # #!/usr/bin/python
# #coding:utf-8
#
# 面试题08.03.魔术索引
#
# https://leetcode-cn.com/problems/magic-index-lcci/
#
# 魔术索引。 在数组A[0...n-1]中，有所谓的魔术索引，满足条件A[i] = i。给定一个有序整数数组，编写一种方法找出魔术索引，若有的话，在数组A中找出一个魔术索引，如果没有，则返回-1。若有多个魔术索引，返回索引值最小的一个。
# 示例1:
#
# 输入：nums = [0, 2, 3, 4, 5]
# 输出：0
# 说明: 0下标的元素为0
#
#
# 示例2:
#
# 输入：n... |
57d4e31c32391b66289da0fe14c29017a35a1973 | wulinlw/leetcode_cn | /初级算法/array_8.py | 1,668 | 3.8125 | 4 | #!/usr/bin/python
#coding:utf-8
# https://leetcode-cn.com/explore/interview/card/top-interview-questions-easy/1/array/28/
# 移动零
# 给定一个数组 nums，编写一个函数将所有 0 移动到数组的末尾，同时保持非零元素的相对顺序。
# 示例:
# 输入: [0,1,0,3,12]
# 输出: [1,3,12,0,0]
# 说明:
# 必须在原数组上操作，不能拷贝额外的数组。
# 尽量减少操作次数。
class Solution(object):
def moveZeroes(self, nums)... |
cc15abee5c3b984f0655422e058e242d235553c3 | wulinlw/leetcode_cn | /leetcode-vscode/817.链表组件.py | 2,469 | 3.671875 | 4 | #
# @lc app=leetcode.cn id=817 lang=python3
#
# [817] 链表组件
#
# https://leetcode-cn.com/problems/linked-list-components/description/
#
# algorithms
# Medium (55.78%)
# Likes: 31
# Dislikes: 0
# Total Accepted: 5.1K
# Total Submissions: 9K
# Testcase Example: '[0,1,2,3]\n[0,1,3]'
#
# 给定一个链表（链表结点包含一个整型值）的头结点 head。
... |
20cff1c9700d9009a225c7413e248a2ee1c48322 | wulinlw/leetcode_cn | /leetcode-vscode/912.排序数组.py | 5,906 | 3.84375 | 4 | #
# @lc app=leetcode.cn id=912 lang=python3
#
# [912] 排序数组
#
# https://leetcode-cn.com/problems/sort-an-array/description/
#
# algorithms
# Medium (53.09%)
# Likes: 68
# Dislikes: 0
# Total Accepted: 29.9K
# Total Submissions: 52.7K
# Testcase Example: '[5,2,3,1]'
#
# 给你一个整数数组 nums，将该数组升序排列。
#
#
#
#
#
#
# ... |
3e400e6a75eb427dc43cce66ff23c5e6bf40a9a2 | wulinlw/leetcode_cn | /初级算法/mathematics_1.py | 1,133 | 3.875 | 4 | #!/usr/bin/python
#coding:utf-8
# https://leetcode-cn.com/explore/interview/card/top-interview-questions-easy/25/math/60/
# Fizz Buzz
# 写一个程序，输出从 1 到 n 数字的字符串表示。
# 1. 如果 n 是3的倍数，输出“Fizz”；
# 2. 如果 n 是5的倍数，输出“Buzz”；
# 3.如果 n 同时是3和5的倍数，输出 “FizzBuzz”。
# 示例：
# n = 15,
# 返回:
# [
# "1",
# "2",
# "Fizz",
# ... |
6719f2cc32c1a8cd9f06575e3c730105c15b3fc5 | wulinlw/leetcode_cn | /字节跳动/array-and-sorting_8.py | 2,455 | 3.890625 | 4 | #!/usr/bin/python
#coding:utf-8
# https://leetcode-cn.com/explore/interview/card/bytedance/243/array-and-sorting/1036/
# 朋友圈
# 班上有 N 名学生。其中有些人是朋友，有些则不是。他们的友谊具有是传递性。如果已知 A 是 B 的朋友，B 是 C 的朋友，那么我们可以认为 A 也是 C 的朋友。所谓的朋友圈，是指所有朋友的集合。
# 给定一个 N * N 的矩阵 M，表示班级中学生之间的朋友关系。如果M[i][j] = 1，表示已知第 i 个和 j 个学生互为朋友关系，否则为不知道。你必须输出所有学生中的已知的... |
177cb9cffdfe5c9fe8035ec10664006982f49606 | wulinlw/leetcode_cn | /初级算法/other_2.py | 1,212 | 4.0625 | 4 | #!/usr/bin/python
#coding:utf-8
# https://leetcode-cn.com/explore/interview/card/top-interview-questions-easy/26/others/65/
# 汉明距离
# 两个整数之间的汉明距离指的是这两个数字对应二进制位不同的位置的数目。
# 给出两个整数 x 和 y，计算它们之间的汉明距离。
# 注意：
# 0 ≤ x, y < 231.
# 示例:
# 输入: x = 1, y = 4
# 输出: 2
# 解释:
# 1 (0 0 0 1)
# 4 (0 1 0 0)
# ↑ ↑
# 上面的箭头指出了对应二... |
c842b9d5b5342e91b9e84dbeecf698e1a8ce8570 | wulinlw/leetcode_cn | /初级算法/mathematics_2.py | 1,763 | 3.984375 | 4 | #!/usr/bin/python
#coding:utf-8
# https://leetcode-cn.com/explore/interview/card/top-interview-questions-easy/25/math/61/
# 计数质数
# 统计所有小于非负整数 n 的质数的数量。
# 示例:
# 输入: 10
# 输出: 4
# 解释: 小于 10 的质数一共有 4 个, 它们是 2, 3, 5, 7 。
# 厄拉多塞筛法
# 西元前250年，希腊数学家厄拉多塞(Eeatosthese)想到了一个非常美妙的质数筛法，
# 减少了逐一检查每个数的的步骤，可以比较简单的从一大堆数字之中，筛选出质数来，这方法... |
8930fdd2af4503f19f5eea56e0e004319275a342 | wulinlw/leetcode_cn | /程序员面试金典/面试题16.16.部分排序.py | 2,042 | 4.03125 | 4 | # #!/usr/bin/python
# #coding:utf-8
#
# 面试题16.16.部分排序
#
# https://leetcode-cn.com/problems/sub-sort-lcci/
#
# 给定一个整数数组，编写一个函数，找出索引m和n，只要将索引区间[m,n]的元素排好序，整个数组就是有序的。注意：n-m尽量最小，也就是说，找出符合条件的最短序列。函数返回值为[m,n]，若不存在这样的m和n（例如整个数组是有序的），请返回[-1,-1]。
# 示例：
# 输入： [1,2,4,7,10,11,7,12,6,7,16,18,19]
# 输出： [3,9]
#
# 提示：
#
# 0
#
#... |
a7a068cb60a7c34934ca3980b45ab240ac077b7e | wulinlw/leetcode_cn | /leetcode-vscode/892.三维形体的表面积.py | 1,718 | 3.671875 | 4 | #
# @lc app=leetcode.cn id=892 lang=python3
#
# [892] 三维形体的表面积
#
# https://leetcode-cn.com/problems/surface-area-of-3d-shapes/description/
#
# algorithms
# Easy (55.73%)
# Likes: 68
# Dislikes: 0
# Total Accepted: 8.6K
# Total Submissions: 14.3K
# Testcase Example: '[[2]]'
#
# 在 N * N 的网格上，我们放置一些 1 * 1 * 1 的立方体... |
2a7878e20f2170d581b6defd02672e1d886cf7e6 | wulinlw/leetcode_cn | /程序员面试金典/面试题02.04.分割链表.py | 1,492 | 3.921875 | 4 | #!/usr/bin/python
#coding:utf-8
# 面试题 02.04. 分割链表
# 编写程序以 x 为基准分割链表，使得所有小于 x 的节点排在大于或等于 x 的节点之前。如果链表中包含 x，x 只需出现在小于 x 的元素之后(如下所示)。
# 分割元素 x 只需处于“右半部分”即可，其不需要被置于左右两部分之间。
# 示例:
# 输入: head = 3->5->8->5->10->2->1, x = 5
# 输出: 3->1->2->10->5->5->8
# https://leetcode-cn.com/problems/partition-list-lcci/
# Definition for... |
ecfa4146a927249cf7cb510dbf14432cd2bb84a7 | wulinlw/leetcode_cn | /剑指offer/30_包含min函数的栈.py | 1,296 | 4.125 | 4 | #!/usr/bin/python
#coding:utf-8
# // 面试题30：包含min函数的栈
# // 题目：定义栈的数据结构，请在该类型中实现一个能够得到栈的最小元素的min
# // 函数。在该栈中，调用min、push及pop的时间复杂度都是O(1)。
class StackWithMin:
def __init__(self):
self.stack = []
self.min_stack = []
def push(self, node):
# write code here
self.stack.append(node)
... |
9e2c9cc88b442a408d79363289c2cfc3905d13c4 | wulinlw/leetcode_cn | /剑指offer/17_打印1到最大的n位数.py | 1,245 | 3.6875 | 4 | #!/usr/bin/python
#coding:utf-8
# 打印1到最大的n位数
# 输入数字n, 按顺序打印从1最大的n位十进制数
# 比如输入3, 则打印出1、2、3、到最大的3位数即999
class Solution:
def Print1ToMaxOfNDigits(self, n):
for i in range(10): #套路写法，生产每一位的0-9，从最左边开始生成
self.recursion(str(i), n, 0) #每个数字的开头
# s 数字开... |
1292292f8f86615a11e933a7234211ad43a71da8 | wulinlw/leetcode_cn | /top-interview-quesitons-in-2018/dynamic-programming_1.py | 1,113 | 3.953125 | 4 | #!/usr/bin/python
#coding:utf-8
# https://leetcode-cn.com/explore/featured/card/top-interview-quesitons-in-2018/272/dynamic-programming/1174/
# 至少有K个重复字符的最长子串
# 找到给定字符串（由小写字符组成）中的最长子串 T ， 要求 T 中的每一字符出现次数都不少于 k 。输出 T 的长度。
# 示例 1:
# 输入:
# s = "aaabb", k = 3
# 输出:
# 3
# 最长子串为 "aaa" ，其中 'a' 重复了 3 次。
# 示例 2:
# 输入:
# s = "a... |
fcc7f51524ae8699c60101b37ff9bbcffdfa1263 | wulinlw/leetcode_cn | /剑指offer/44_数字序列中某一位的数字.py | 2,798 | 3.75 | 4 | #!/usr/bin/python
#coding:utf-8
# // 面试题44：数字序列中某一位的数字
# // 题目：数字以0123456789101112131415…的格式序列化到一个字符序列中。在这
# // 个序列中，第5位（从0开始计数）是5，第13位是1，第19位是4，等等。请写一
# // 个函数求任意位对应的数字。
class Solution:
def digitAtIndex(self, index):
if index < 0:
return -1
digits = 1
while True:
le... |
93980a2f1b9d778ff907998b6fb722722ec28d73 | wulinlw/leetcode_cn | /递归/recursion_1_1.py | 1,304 | 4.15625 | 4 | #!/usr/bin/python
#coding:utf-8
# https://leetcode-cn.com/explore/orignial/card/recursion-i/256/principle-of-recursion/1198/
# 反转字符串
# 编写一个函数，其作用是将输入的字符串反转过来。输入字符串以字符数组 char[] 的形式给出。
# 不要给另外的数组分配额外的空间，你必须原地修改输入数组、使用 O(1) 的额外空间解决这一问题。
# 你可以假设数组中的所有字符都是 ASCII 码表中的可打印字符。
# 示例 1：
# 输入：["h","e","l","l","o"]
# 输出：["o","l",... |
b3328bb716cac18bf8e375f48de6ae5d67faa44b | wulinlw/leetcode_cn | /中级算法/tree_6.py | 2,059 | 3.796875 | 4 | #!/usr/bin/python
#coding:utf-8
# https://leetcode-cn.com/explore/interview/card/top-interview-questions-medium/32/trees-and-graphs/90/
# 岛屿的个数
# 给定一个由 '1'（陆地）和 '0'（水）组成的的二维网格，计算岛屿的数量。一个岛被水包围，并且它是通过水平方向或垂直方向上相邻的陆地连接而成的。你可以假设网格的四个边均被水包围。
# 示例 1:
# 输入:
# 11110
# 11010
# 11000
# 00000
# 输出: 1
# 示例 2:
# 输入:
# 11000
# ... |
bacfbc3a4a068cf87954be2a53e0a6ab44ba41bc | wulinlw/leetcode_cn | /链表/linked-list_5_3.py | 2,469 | 4.125 | 4 | #!/usr/bin/python
# coding:utf-8
# https://leetcode-cn.com/explore/learn/card/linked-list/197/conclusion/764/
# 扁平化多级双向链表
# 您将获得一个双向链表，除了下一个和前一个指针之外，它还有一个子指针，可能指向单独的双向链表。这些子列表可能有一个或多个自己的子项，依此类推，生成多级数据结构，如下面的示例所示。
# 扁平化列表，使所有结点出现在单级双链表中。您将获得列表第一级的头部。
# 示例:
# 输入:
# 1---2---3---4---5---6--NULL
# |
# ... |
43d875814e422cab3a1d28b38b8862ef137e70ae | wulinlw/leetcode_cn | /剑指offer/28_对称的二叉树.py | 2,014 | 3.671875 | 4 | #!/usr/bin/python
#coding:utf-8
# // 面试题28：对称的二叉树
# // 题目：请实现一个函数，用来判断一棵二叉树是不是对称的。如果一棵二叉树和
# // 它的镜像一样，那么它是对称的。
class TreeNode:
def __init__(self, x):
self.val = x
self.left = None
self.right = None
class Solution:
def isSymmetrical(self, root):
if not root:return True
... |
b7fce39adf1ba1fa6029dda276e2fde9eb8277ec | wulinlw/leetcode_cn | /剑指offer/23_链表中环的入口结点.py | 1,404 | 3.796875 | 4 | #!/usr/bin/python
#coding:utf-8
# // 面试题23：链表中环的入口结点
# // 题目：一个链表中包含环，如何找出环的入口结点？例如，在图3.8的链表中，
# // 环的入口结点是结点3。
class ListNode:
def __init__(self, x=None):
self.val = x
self.next = None
class Solution:
def initlinklist(self, nums):
head = ListNode(nums[0])
re = head
for... |
4154a18778d1e344a20d388bb08ce7d33022adce | wulinlw/leetcode_cn | /leetcode-vscode/78.子集.py | 1,086 | 3.578125 | 4 | #
# @lc app=leetcode.cn id=78 lang=python3
#
# [78] 子集
#
# https://leetcode-cn.com/problems/subsets/description/
#
# algorithms
# Medium (76.55%)
# Likes: 493
# Dislikes: 0
# Total Accepted: 69.2K
# Total Submissions: 90.1K
# Testcase Example: '[1,2,3]'
#
# 给定一组不含重复元素的整数数组 nums，返回该数组所有可能的子集（幂集）。
#
# 说明：解集不能包含重复... |
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