blob_id stringlengths 40 40 | repo_name stringlengths 5 127 | path stringlengths 2 523 | length_bytes int64 22 545k | score float64 3.5 5.34 | int_score int64 4 5 | text stringlengths 22 545k |
|---|---|---|---|---|---|---|
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... |
80082f949d1589ddc0a79a9067b95413e873b4f4 | wulinlw/leetcode_cn | /leetcode-vscode/406.根据身高重建队列.py | 1,933 | 3.546875 | 4 | #
# @lc app=leetcode.cn id=406 lang=python3
#
# [406] 根据身高重建队列
#
# https://leetcode-cn.com/problems/queue-reconstruction-by-height/description/
#
# algorithms
# Medium (62.74%)
# Likes: 272
# Dislikes: 0
# Total Accepted: 21.8K
# Total Submissions: 34.2K
# Testcase Example: '[[7,0],[4,4],[7,1],[5,0],[6,1],[5,2]]... |
e727d73d45d67bf07c8a7cf7bcf3103d8acd75b4 | wulinlw/leetcode_cn | /数组和字符串/array_6.py | 939 | 3.75 | 4 | #!/usr/bin/python
#coding:utf-8
# https://leetcode-cn.com/explore/featured/card/array-and-string/202/conclusion/792/
# 杨辉三角 II
# 给定一个非负索引 k，其中 k ≤ 33，返回杨辉三角的第 k 行。
# 在杨辉三角中，每个数是它左上方和右上方的数的和。
# 示例:
# 输入: 3
# 输出: [1,3,3,1]
# 进阶：
# 你可以优化你的算法到 O(k) 空间复杂度吗？
class Solution(object):
def getRow(self, rowIndex):
"... |
e17240c468f25e13762598f4f7c18bb9fbacd888 | wulinlw/leetcode_cn | /二叉树/search-tree_1_2.py | 1,994 | 4.09375 | 4 | #!/usr/bin/python
#coding:utf-8
# https://leetcode-cn.com/explore/learn/card/introduction-to-data-structure-binary-search-tree/64/introduction-to-a-bst/172/
# 二叉搜索树迭代器
# 实现一个二叉搜索树迭代器。你将使用二叉搜索树的根节点初始化迭代器。
# 调用 next() 将返回二叉搜索树中的下一个最小的数。
# 示例：
# BSTIterator iterator = new BSTIterator(root);
# iterator.next(); // 返回 ... |
1926f0d51153da212fbfd132588b7547ca9b9e9d | wulinlw/leetcode_cn | /初级算法/linkedList_4.py | 1,718 | 4.25 | 4 | #!/usr/bin/python
#coding:utf-8
# https://leetcode-cn.com/explore/interview/card/top-interview-questions-easy/6/linked-list/44/
# 合并两个有序链表
# 将两个有序链表合并为一个新的有序链表并返回。新链表是通过拼接给定的两个链表的所有节点组成的。
# 示例：
# 输入：1->2->4, 1->3->4
# 输出：1->1->2->3->4->4
# 新建链表，对比两个链表指针，小的放新链表中，直到某条链表结束，
# 将另一条链表剩余部分接入新链表
# Definition for singly-l... |
1270af1e1c684481ba5488a2fcff3aa3ca656284 | wulinlw/leetcode_cn | /leetcode-vscode/637.二叉树的层平均值.py | 1,670 | 3.703125 | 4 | #
# @lc app=leetcode.cn id=637 lang=python3
#
# [637] 二叉树的层平均值
#
# https://leetcode-cn.com/problems/average-of-levels-in-binary-tree/description/
#
# algorithms
# Easy (63.16%)
# Likes: 98
# Dislikes: 0
# Total Accepted: 14.5K
# Total Submissions: 22.9K
# Testcase Example: '[3,9,20,15,7]'
#
# 给定一个非空二叉树, 返回一个由每层节... |
096fdd8e7212763ede543a13cdad2240dd9f91ae | wulinlw/leetcode_cn | /dynamic-programming/dynamic-programming_63.py | 2,054 | 3.75 | 4 | #!/usr/bin/python
#coding:utf-8
# 63. 不同路径 II
# 一个机器人位于一个 m x n 网格的左上角 （起始点在下图中标记为“Start” ）。
# 机器人每次只能向下或者向右移动一步。机器人试图达到网格的右下角（在下图中标记为“Finish”）。
# 现在考虑网格中有障碍物。那么从左上角到右下角将会有多少条不同的路径？
# 网格中的障碍物和空位置分别用 1 和 0 来表示。
# 说明：m 和 n 的值均不超过 100。
# 示例 1
# 输入:
# [
# [0,0,0],
# [0,1,0],
# [0,0,0]
# ]
# 输出: 2
# 解释:
# 3x3 网格的正中间... |
ecaa303c54ccfc6abc2d3e1c76799df9dabe7794 | wulinlw/leetcode_cn | /程序员面试金典/面试题04.01.节点间通路.py | 2,549 | 3.765625 | 4 | # #!/usr/bin/python
# #coding:utf-8
#
# 面试题04.01.节点间通路
#
# https://leetcode-cn.com/problems/route-between-nodes-lcci/
#
# 节点间通路。给定有向图，设计一个算法，找出两个节点之间是否存在一条路径。
# 示例1:
#
# 输入：n = 3, graph = [[0, 1], [0, 2], [1, 2], [1, 2]], start = 0, target = 2
# 输出：true
#
#
# 示例2:
#
# 输入：n = 5, graph = [[0, 1], [0, 2], [0, 4]... |
1ffbfbbbc0ea0c6f4725b89363464d734a17cd94 | wulinlw/leetcode_cn | /剑指offer/3_数组中重复的数字.py | 935 | 3.8125 | 4 | #!/usr/bin/python
#coding:utf-8
# 数组中重复的数字
# 在一个长度为n的数组里的所有数字都在0到n-1的范围内。
# 数组中某些数字是重复的，但不知道有几个数字是重复的。也不知道每个数字重复几次。
# 请找出数组中任意一个重复的数字。
# 例如，如果输入长度为7的数组{2,3,1,0,2,5,3}，那么对应的输出是重复的数字2或者3。
class Solution(object):
def duplicate(self, nums):
if len(nums)== 0:
return False
re = []
for... |
bb2dd47c253fd1d7226b49f6a73d0f93658334c8 | wulinlw/leetcode_cn | /top-interview-quesitons-in-2018/string_3.py | 1,781 | 3.65625 | 4 | #!/usr/bin/python
#coding:utf-8
# https://leetcode-cn.com/explore/featured/card/top-interview-quesitons-in-2018/275/string/1138/
# 单词拆分
# 给定一个非空字符串 s 和一个包含非空单词列表的字典 wordDict，判定 s 是否可以被空格拆分为一个或多个在字典中出现的单词。
# 说明：
# 拆分时可以重复使用字典中的单词。
# 你可以假设字典中没有重复的单词。
# 示例 1：
# 输入: s = "leetcode", wordDict = ["leet", "code"]
# 输出: true... |
7799a49165b44fdc5f7c6064432c6204f384286b | wulinlw/leetcode_cn | /leetcode-vscode/563.二叉树的坡度.py | 1,736 | 3.875 | 4 | #
# @lc app=leetcode.cn id=563 lang=python3
#
# [563] 二叉树的坡度
#
# https://leetcode-cn.com/problems/binary-tree-tilt/description/
#
# algorithms
# Easy (53.09%)
# Likes: 54
# Dislikes: 0
# Total Accepted: 8.9K
# Total Submissions: 16.7K
# Testcase Example: '[1,2,3]'
#
# 给定一个二叉树，计算整个树的坡度。
#
# 一个树的节点的坡度定义即为，该节点左子树的... |
66fbee2c03751ac6ca76c7b9e1b1449cc609597d | wulinlw/leetcode_cn | /leetcode-vscode/98.验证二叉搜索树.py | 4,208 | 3.8125 | 4 | #
# @lc app=leetcode.cn id=98 lang=python3
#
# [98] 验证二叉搜索树
#
# https://leetcode-cn.com/problems/validate-binary-search-tree/description/
#
# algorithms
# Medium (29.10%)
# Likes: 543
# Dislikes: 0
# Total Accepted: 103.5K
# Total Submissions: 340.3K
# Testcase Example: '[2,1,3]'
#
# 给定一个二叉树，判断其是否是一个有效的二叉搜索树。
# ... |
d8ee8504bbeea0dc8d222d93e9a13c16ff5ca4d8 | wulinlw/leetcode_cn | /leetcode-vscode/820.单词的压缩编码.py | 4,185 | 3.796875 | 4 | #
# @lc app=leetcode.cn id=820 lang=python3
#
# [820] 单词的压缩编码
#
# https://leetcode-cn.com/problems/short-encoding-of-words/description/
#
# algorithms
# Medium (40.03%)
# Likes: 70
# Dislikes: 0
# Total Accepted: 14.1K
# Total Submissions: 32.8K
# Testcase Example: '["time", "me", "bell"]'
#
# 给定一个单词列表，我们将这个列表编码... |
15acb129012387c2a2c05e982a16a9e2d454660a | wulinlw/leetcode_cn | /leetcode-vscode/897.递增顺序查找树.py | 2,420 | 3.9375 | 4 | #
# @lc app=leetcode.cn id=897 lang=python3
#
# [897] 递增顺序查找树
#
# https://leetcode-cn.com/problems/increasing-order-search-tree/description/
#
# algorithms
# Easy (65.76%)
# Likes: 58
# Dislikes: 0
# Total Accepted: 8K
# Total Submissions: 12.1K
# Testcase Example: '[5,3,6,2,4,null,8,1,null,null,null,7,9]'
#
# 给... |
3192958328b62328c5bcc2c89a2ef2fc53ba001b | wulinlw/leetcode_cn | /leetcode-vscode/606.根据二叉树创建字符串.py | 2,203 | 3.640625 | 4 | #
# @lc app=leetcode.cn id=606 lang=python3
#
# [606] 根据二叉树创建字符串
#
# https://leetcode-cn.com/problems/construct-string-from-binary-tree/description/
#
# algorithms
# Easy (52.76%)
# Likes: 98
# Dislikes: 0
# Total Accepted: 9.4K
# Total Submissions: 17.7K
# Testcase Example: '[1,2,3,4]'
#
# 你需要采用前序遍历的方式，将一个二叉树转换... |
940b4d34f8c3c69f861cb1073cb9935f503835bf | wulinlw/leetcode_cn | /leetcode-vscode/350.两个数组的交集-ii.py | 1,630 | 3.765625 | 4 | #
# @lc app=leetcode.cn id=350 lang=python3
#
# [350] 两个数组的交集 II
#
# https://leetcode-cn.com/problems/intersection-of-two-arrays-ii/description/
#
# algorithms
# Easy (48.66%)
# Likes: 315
# Dislikes: 0
# Total Accepted: 103.1K
# Total Submissions: 205.9K
# Testcase Example: '[1,2,2,1]\n[2,2]'
#
# 给定两个数组，编写一个函数来... |
2d2c4fe3cd12f51c41aeb1dea6920d4d2752bb53 | wulinlw/leetcode_cn | /剑指offer/43_从1到n整数中1出现的次数.py | 3,313 | 3.640625 | 4 | #!/usr/bin/python
#coding:utf-8
# // 面试题43：从1到n整数中1出现的次数
# // 题目：输入一个整数n，求从1到n这n个整数的十进制表示中1出现的次数。例如
# // 输入12，从1到12这些整数中包含1 的数字有1，10，11和12，1一共出现了5次。
class Solution:
def NumberOf1Between1AndN_Solution(self, n):
# https://blog.csdn.net/ggdhs/article/details/90311852
# 如果要计算百位上1出现的次数，它要受到3方面的影响：百位上的数... |
db50eae1594cc0bc7ee8f331ca3b6383a3217288 | wulinlw/leetcode_cn | /递归/recursion_5_3.py | 4,062 | 3.9375 | 4 | #!/usr/bin/python
#coding:utf-8
# https://leetcode-cn.com/explore/orignial/card/recursion-i/260/conclusion/1233/
# 不同的二叉搜索树 II
# 给定一个整数 n，生成所有由 1 ... n 为节点所组成的二叉搜索树。
# 示例:
# 输入: 3
# 输出:
# [
# [1,null,3,2],
# [3,2,null,1],
# [3,1,null,null,2],
# [2,1,3],
# [1,null,2,null,3]
# ]
# 解释:
# 以上的输出对应以下 5 种不同结构的二叉搜... |
b43fa3334180039300923931d9259fe5a6ba832e | wulinlw/leetcode_cn | /leetcode-vscode/69.x-的平方根.py | 1,050 | 3.671875 | 4 | #
# @lc app=leetcode.cn id=69 lang=python3
#
# [69] x 的平方根
#
# https://leetcode-cn.com/problems/sqrtx/description/
#
# algorithms
# Easy (37.57%)
# Likes: 364
# Dislikes: 0
# Total Accepted: 126.8K
# Total Submissions: 335.5K
# Testcase Example: '4'
#
# 实现 int sqrt(int x) 函数。
#
# 计算并返回 x 的平方根，其中 x 是非负整数。
#
# 由... |
72aadd6ad7c82acd94fd6500aea2f08d406da84e | wulinlw/leetcode_cn | /哈希表/hash-table_1_1.py | 2,228 | 3.984375 | 4 | #!/usr/bin/python
# coding:utf-8
# https://leetcode-cn.com/explore/learn/card/hash-table/203/design-a-hash-table/799/
# 设计哈希集合
# 不使用任何内建的哈希表库设计一个哈希集合
# 具体地说，你的设计应该包含以下的功能
# add(value)：向哈希集合中插入一个值。
# contains(value) ：返回哈希集合中是否存在这个值。
# remove(value)：将给定值从哈希集合中删除。如果哈希集合中没有这个值，什么也不做。
# 示例:
# MyHashSet hashSet = new My... |
54cd0bace7d37beaf42ea3520896257a5bd3d19b | wulinlw/leetcode_cn | /中级算法/sorting-and-searching_5.py | 2,508 | 3.765625 | 4 | #!/usr/bin/python
#coding:utf-8
# https://leetcode-cn.com/explore/interview/card/top-interview-questions-medium/50/sorting-and-searching/100/
# 在排序数组中查找元素的第一个和最后一个位置
# 给定一个按照升序排列的整数数组 nums，和一个目标值 target。找出给定目标值在数组中的开始位置和结束位置。
# 你的算法时间复杂度必须是 O(log n) 级别。
# 如果数组中不存在目标值，返回 [-1, -1]。
# 示例 1:
# 输入: nums = [5,7,7,8,8,10],... |
5f492700d7482fa2df271ec9088f8f06fbeeb568 | Bhavya-Tripathi/ComputerNetworks | /Assignment-3/test.py | 1,245 | 3.765625 | 4 | import socket,sys
print("Setting up server: ")
#Get hostname,IP
sock = socket.socket()
hostname = socket.gethostname()
ip = socket.gethostbyname(hostname)
port = 6969
sock.bind((hostname,port)) #binds IP of localhost to port
print(hostname, '({})'.format(ip))
name = input('Enter your name:')
sock.listen(1) #to locate... |
efa5b93f22db52d9dfe98f8a95f3c8da4866aa24 | VictorSHJ/fundamentos-python | /palindroma.py | 1,116 | 4.125 | 4 | # GRUPO 2:
# Crea una funcion que dado una palabra diga si es palindroma o no.
def palindroma(palabra):
print(f"Palabra Normal: {palabra}, Palabra Invertida: {palabra[::-1]}")
if palabra == palabra[::-1]:
print("Es palindroma")
else:
print("No es palindroma")
print("Ingrese una palabra :"... |
893ad93e796a1e7ac5fd1df23c8ab4a4c432a454 | SebastianStaab/AEMLproject | /tetris/testing.py | 806 | 3.515625 | 4 | cols = 10
rows = 20
board = [
[ 0 for x in range(cols) ]
for y in range(rows)
]
board += [[ 1 for x in range(cols)]]
def join_matrixes(mat1, mat2, mat2_off):
off_x, off_y = mat2_off
for cy, row in enumerate(mat2):
for cx, val in enumerate(row):
mat1[cy+off_y-1 ][cx+off_... |
b7908c0d09dd1a60a6c355b67ad4d7f2b5b324a6 | jamesdschmidt/exercises-for-programmers | /23-troubleshooting-car-issues/troubleshooting_car_issues.py | 1,024 | 3.984375 | 4 | answer = input("Is the car silent when you turn the key? ")
if answer.lower().startswith("y"):
answer = input("Are the battery terminals corroded? ")
if answer.lower().startswith("y"):
print("Clean terminals and try starting again.")
else:
print("Replace cables and try again.")
else:
ans... |
df0d8ad8924c8948c854619cef38c87f90a74e7e | jamesdschmidt/exercises-for-programmers | /11-currency-conversion/currency_conversion.py | 309 | 3.875 | 4 | DOLLAR_RATE = 100
euros = int(input("How many euros are you exchanging? "))
exchange_rate = float(input("What is the exchange rate? "))
dollars = round((euros * exchange_rate) / DOLLAR_RATE, 2)
print(f"{euros} euros at an exchange rate of {exchange_rate:,.2f} is\n"
f"{dollars:,.2f} U.S. dollars.")
|
a81d2c7a6c3213eb61126ec23759d462e0b6862c | jamesdschmidt/exercises-for-programmers | /14-tax-calculator/tax_calculator.py | 305 | 3.75 | 4 | TAX_RATE = 0.055
amount = float(input("What is the order amount? "))
state = input("What is the state? ")
total = amount
output = ""
if "wi" == state.lower():
tax = amount * TAX_RATE
output = f"The tax is ${tax:,.2f}.\n"
total += tax
output += f"The total is ${total:,.2f}"
print(output)
|
e75d330b5e9e8fa368266108785fad07875ffd97 | jamesdschmidt/exercises-for-programmers | /16-legal-driving-age/legal_driving_age.py | 119 | 4.09375 | 4 | age = int(input("What is your age? "))
print("You", "are not" if age < 16 else "are", "old enough to legally drive.")
|
bbf1b7caf79ccf7f09fa42a45915a12feb88eb8d | jamesdschmidt/exercises-for-programmers | /13-determining-compound-interest/determining_compound_interest.py | 503 | 3.734375 | 4 | principal = float(input("What is the principal amount? "))
rate = float(input("What is the rate? "))
years = int(input("What is the number of years? "))
compound = int(input("What is the number of times the interest\n"
"is compounded per year?"))
amount = round(principal * ((1 + (rate / 100) / com... |
14a129dab9189feff13e5be95e38aae5fefb5b36 | dwighthubbard/ipython_embedded_notebooks | /intro/3_tail.py | 754 | 3.828125 | 4 | # Blink an LED on pin 18.
# Connect a low-ohm (like 360 ohm) resistor in series with the LED.
import RPi.GPIO as GPIO
import time
# A variable so we can change the PIN number for this script in once place
# if we move the LED to a different pin.
PIN = 7
# Set the pin to do output
GPIO.setmode(GPIO.BCM)
GPIO.setup(PI... |
b107480b098897c85a77f5bfa88ea20a4f78a497 | AlNiLo/LearnPython | /HomeW/!!!!L2_test_classroom_list.py | 1,326 | 3.875 | 4 | quest = input('По какому классу вывести оценки? ')
school = [
{'class':[
{'1':[
{'a':[1, 3, 4, 4, 3, 5, 3],
'b':[1, 2, 4, 4, 3, 4, 4],
'c':[1, 2, 2, 5, 4, 5, 5]
}
],
... |
e5f82bbb3538305b49f22f9c1e8530a0e87a20fa | kardeepak/protostar | /format0.py | 228 | 3.609375 | 4 | import struct
# padding will pad and print an integer fron the stack. It will print 64 integers
padding = "%64d"
# Value that will be written to the target variable
value = struct.pack("I", 0xdeadbeef)
print(padding + value)
|
498cdb043ed9f878303fedb8d72019ce27d8faa0 | CodecoolBP20173/game_enhancement | /tictac_test.py | 6,627 | 3.71875 | 4 | import string
field = []
player_x_win = 0
player_o_win = 0
def print_field(): # print empty field with nested lists
alphabet = string.ascii_lowercase
print("\n" * 50)
for i in range(field_size):
field.append([" "])
for j in range(field_size):
field[i].append(" ")
... |
cfba2c4ac2beb767417b1e8e46486dea23cdf26e | adamcharnock/repose | /repose/validators.py | 1,626 | 3.703125 | 4 | from booby.validators import *
import six
class Range(Integer):
"""This validator forces fields values to be within a given range (inclusive)"""
def __init__(self, min=None, max=None):
self.min = min
self.max = max
@nullable
def validate(self, value):
if self.max is not None:... |
5a0be252fff51c358f1360716f892c69078c98a1 | mirzaevolution/Python-Basics | /PythonApplication2/Statements.py | 790 | 3.890625 | 4 | import constant
"""
Python Program #1
"""
#line continuation statement
number1 = 1 + 2 + \
3 + 4 + \
5 + 6
number2 = (1 + 2 +
3 + 4 +
5 + 6)
print(number1)
print(number2)
# One line assignment
var1,var2,var3 = 1,12.4,"Mirza Ghulam Rasyid"
print(var1)
print(var2)
print... |
ca5941b7b2355135eb03c3d8880a741db9a221f2 | November29th/ObjectedOrientedCards | /objectOrientedCards.py | 1,719 | 3.796875 | 4 | from random import*
class Card(object):
def __init__(self, suit = None, value = None):
self.value = value
class Diamond(Card):
def __init__(self):
super(Diamond, self).__init__("Diamond", 4)
class Heart(Card):
def __init__(self):
super(Heart, self).__init__("Heart", 3)
class Club... |
20bb2a14fbb695fa1d1868147e8e2afc147cecc3 | fatychang/pyimageresearch_examples | /ch10 Neural Network Basics/perceptron_example.py | 1,157 | 4.25 | 4 | # -*- coding: utf-8 -*-
"""
Created on Mon Dec 9 12:56:47 2019
This is an example for runing percenptron structure to predict bitwise dataset
You may use AND, OR and XOR in as the dataset.
A preceptron class is called in this example.
An example from book deep learning for computer vision with Python ch10
@author: ... |
fd608e74e03f957482ac06f6b55b9f2f3363434a | fatychang/pyimageresearch_examples | /barcode-detection-guide/detect_barcode.py | 2,746 | 3.578125 | 4 | # -*- coding: utf-8 -*-
"""
Created on Tue Feb 4 13:04:46 2020
This script demonstrates the technique to detect the barcode
using computer vision and image processing technique
A edge detection approach is implemented in this script with opencv
This sample is inspired by the post from pyinagesearch
@author: jscha... |
b7c3cf717dea0f7c9b95599c99e691e6b0fe16ea | amcatlin/CSCI156-ExceptionHandling | /ExceptionHandling answers.py | 1,155 | 3.859375 | 4 | __author__ = 'Alicia'
y = 'Enter SS#: '
inputstring = 'Please enter a valid social security number of the form ###-##-#### including the dashes: '
def question(s):
social = input(s).strip()
try:
AAA, GG, SSSS = social.split('-')
area = int(AAA)
group = int(GG)
serial = int(SSSS... |
84f18c6f127360133af6802d2cd698551d6a6665 | rarafa/crypto-course | /rot13.py | 778 | 3.765625 | 4 | #!/usr/bin/python3
import sys
def rot13(input):
output=""
for i in input:
if i.isalpha():
if i.islower():
if (ord(i)+13)%ord('z') < ord('a'):
output += chr( ((ord(i)+13)%ord('z'))+ord('a') )
else:
output += chr( (ord(i... |
c83a41d51e6061033feaf6054f63c3f570c1f03b | junejunejune/numpy_practice | /sort.py | 460 | 3.671875 | 4 | import numpy as np
def selection_sort(x):
for i in range(len(x)):
swap = i + np.argmin(x[i:])
(x[i], x[swap]) = (x[swap], x[i])
return x
def bogosort(x):
while np.any(x[:-1] > x[1:]):
np.random.shuffle(x)
return x
x = np.array([2,1,4,3,5])
print(selection_sort(x))
x = np.arra... |
b7a04a3f5513cfa9d6aca9f3cb548726ec953326 | Hegerstrand/energyMapperOld | /Python/venv/Lib/site-packages/copyfile/copyfile.py | 1,403 | 3.625 | 4 | import os
import shutil
import logging
def touch(fname, times=None):
"""Creates an empty file at fname, creating path if necessary
Answer taken from Stack Overflow http://stackoverflow.com/a/1160227
User: ephemient http://stackoverflow.com/users/20713
License: CC-BY-SA 3.0 https://creativecommons.org/... |
f2e5accccddc156ccc92d48cca4b7bc2a1f50d4f | szwagiers/sqlite | /ccsqlite/lesson3.py | 319 | 3.96875 | 4 | import sqlite3
conn = sqlite3.connect("database.db")
# put cursor
c = conn.cursor()
# execute sql directly to database
c.execute("CREATE TABLE IF NOT EXISTS books (title TEXT,pages INTEGER)")
c.execute('INSERT INTO books VALUES("The Count of Monte Cristo",1316)')
conn.commit()
c.execute('SELECT * FROM books')
|
2423751e0607aa434d00d5e909e19126ac366ee9 | PPCCCSCS/Flippant | /FlippantTutor.py | 91,171 | 4.28125 | 4 | """
File: FlippantTutor.py
The program teaches users how to play the dice game Flippant
via a simulated game with one to three computer-controlled
opponents. Their choices are chosen randomly, so players
shouldn't expect to learn strategies for competing with skilled
players, but this program should provide an ... |
feea22a9815e0067f55718d5c59f90f16a5c4c69 | hannahmok/webhax_samples | /04-shellsanitization-better/index.py | 810 | 3.515625 | 4 | #!/usr/bin/env python3
import os
import sys
import urllib.parse
# parse the query string (?x=1&y=2) into the parameters dict {'x': ['1'], 'y': ['2']}
parameters = urllib.parse.parse_qs(os.environ['QUERY_STRING'])
# handle ?query=xx to search for text
if 'query' in parameters:
print('Content-type: text/plain\n')
... |
f963c5c6e26976b826456b634c80ed388d7cde55 | ptsteadman/qfo-algo-console | /strategy/monkey.py | 370 | 3.6875 | 4 | import random
class Monkey(object):
""" Silly monkey strategy every freq tick roll the dice
To instantiate a 30 tick monkey:
>>> monkey = Monkey(30)
"""
def __init__(self, freq):
self.freq = freq
def __call__(self, tick):
if tick.index % self.freq:
return None
... |
a73e83cd06a469cbbea8440e836f359715d1ff08 | ptsteadman/qfo-algo-console | /strategy/bollinger.py | 920 | 3.53125 | 4 | class Bollinger(object):
""" Bollinger's band trading strategy, for Mean Reversion or Breakout
Bollinger's band take 2 parameters the period N of the underlying moving
average and the widht of the band K.
To instantiate a 20 days, 2 standard dev:
>>> bollinger = Bollinger(20, 2, reversion)
""... |
847ace6bebef81ef053d6a0268fa54e36072dd72 | chenshaobin/python_100 | /ex2.py | 1,113 | 4.1875 | 4 | #!/usr/bin/python
# -*- coding: utf-8 -*-
"""
# Write a program which can compute the factorial of a given numbers.
# The results should be printed in a comma-separated sequence on a single line.Suppose the following input is supplied to the program: 8 Then, the output should be:40320
"""
# 使用while
"""
n = int... |
baa5ff5f08103e624b90c7a754f0f5cc60429f0e | chenshaobin/python_100 | /ex9.py | 581 | 4.15625 | 4 | #!/usr/bin/python
# -*- coding: utf-8 -*-
"""
# Write a program that accepts sequence of lines as input
# and prints the lines after making all characters in the sentence capitalized.
"""
# solution1
"""
lst = []
while True:
x = input("Please enter one word:")
if len(x) == 0:
break
lst.appen... |
04d0d49313ad3fd020a1dd5364f663819160c825 | wongxinjie/python-tools | /profile/timer.py | 1,177 | 3.5625 | 4 | #!/usr/bin/env python3
# -*- coding: utf-8 -*-
"""程序片函数运行时间测量模块
usage:
from timer import Timer
...
with Timer():
...
"""
import gc
import timeit
class Timer(object):
def __init__(self, timer=None, disable_gc=False,
verbose=False, program=None):
if timer is None:
... |
fd3a3adc3b657f5f3d72697827658c1ecf804ea9 | seanxwh/github | /Python/miscellaneous/Nerual_Network.py | 5,830 | 4.09375 | 4 | #A neural network implementation for data/labels classification using Python. User can train the network
#by provide data that has labels in it. User can also fine tune the network by modifying it parameters (e.ghidden layers, learning rate, etc)
# Package imports
import matplotlib.pyplot as plt
import math
import num... |
8c551099820670650129116ba923151ccffcc6dc | Agugu95/AlgoPy | /algo_py/BridgeOnTrucks.py | 652 | 3.78125 | 4 | from collections import deque
def solution(bridge_length, weight, truck_weights):
answer = 0
waited_truck = deque(truck_weights)
cur_bridge = [0] * bridge_length
cur_weight = 0
while cur_bridge:
cur_weight -= cur_bridge.pop(0)
if waited_truck:
if(cur_weight + waited_tr... |
ab46a3b79cb7eeeba170fd2db174876fca14b900 | ritik1234/calculator | /calculatorgui.py | 5,091 | 3.71875 | 4 | from tkinter import *
import math
root = Tk()
def click(event):
global value
text = event.widget.cget("text")
if text == "sin":
value.set(math.sin(float(exp.get())))
exp.update()
elif text == "cos":
value.set(math.cos(float(exp.get())))
exp.update()
elif t... |
855fab989e0c28a10cfd7459bfcf6cd03091a061 | AlibekAbd/- | /Gauss.py | 826 | 3.71875 | 4 | import numpy as np
import random
import scipy.linalg as sla
n = int(input())
A = np.random.rand(n,n)
f = np.random.rand(n)
#diagonal dimension
for i in range(0,n):
summa = 0
for j in range(0,n):
summa += abs(A[i][j])
c = random.uniform(1,2)
A[i][i] = summa + c
def forward_elimination(A, f, n):
for k in range(... |
932dffef241b7aadab1551b63db501e7146457c0 | Ankirama/Epitech | /B5---Java-I-Programming/JWeb/data/initDB.py | 5,787 | 4.09375 | 4 | import sqlite3
import sys
class Database:
'''
Helper to use sqlite3 and init our database
'''
def __init__(self, dbname):
'''
It will connect to the database name (or create it) and create (if not exists) our tables
@param: dbname: database name
'''
try:
... |
0a0e107f75b9549da0db80273d81ff7b4e84e88a | alanmanderson/advent_of_code_2020 | /2/validate_passwords.py | 1,435 | 3.515625 | 4 | INPUT_FILE = 'my_input.txt'
def read_file(filename):
with open(filename, 'r') as fp:
data = []
for line in fp:
data.append(line.strip().split(': '))
return data
def old_valid_passwords(data):
valid_passwords = []
for str_policy, password in data:
policy = parse_poli... |
eaeed21766f75657270303ddac34c8dcae8f4f01 | Scientific-Computing-at-Temple-Physics/prime-number-finder-gt8mar | /Forst_prime.py | 605 | 4.375 | 4 | # Marcus Forst
# Scientific Computing I
# Prime Number Selector
# This function prints all of the prime numbers between two entered values.
import math as ma
# These functions ask for the number range, and assign them to 'x1' and 'x2'
x1 = int(input('smallest number to check: '))
x2 = int(input('largest number to ... |
d40172caa6c5f944dd0db4a8a2b7af65ec861c63 | piotrbla/pyExamples | /format_poem.py | 580 | 3.78125 | 4 | from unittest.test.test_case import Test
def format_poem(poem):
if poem[-1:] == ".":
return ".\n".join(x.strip() for x in str.split(poem, '.'))
else:
return ".\n".join(x.strip() for x in str.split(poem, '.'))[:-1] + poem[-1:]
x=[1, 2, 3]
a = sum(x)/len(x)
print(format_poem('Beautiful is bette... |
c38197e38f25c0a4897daf349529adb8371fd860 | piotrbla/pyExamples | /chess.py | 578 | 3.609375 | 4 | class Field:
def __init__(self, r, c):
self.r = r
self.c = chr(c + ord('A') - 1)
class Board:
def __init__(self):
self.fields = []
for row in range(1, 9):
row_fields = []
for column in range(1, 9):
row_fields.append(Field(row, column))
... |
eb2dcb63ce695d4787b4269754005b9ff9074834 | piotrbla/pyExamples | /random_stack.py | 4,073 | 3.515625 | 4 | from random import randint
from random import seed
class element:
def __init__(self, data, prev=None):
self.data = data
self.prev = prev
class stos:
def __init__(self, first):
self.length = 1
self.porownania = 0
self.przypisania = 4
self.last = element(first)
... |
7a04c5413d26c1daf88792f0353f5a5ed0872a98 | piotrbla/pyExamples | /resistor.py | 3,887 | 3.734375 | 4 | import unittest
def encode_resistor_colors(ohms_string):
result = ""
codes = {0: "black", 1: "brown", 2: "red", 3: "orange", 4: "yellow", 5: "green", 6: "blue", 7: "violet", 8: "gray",
9: "white"}
without_ohms = ohms_string.replace(" ohms", "")
x = 0
if without_ohms.endswith('k'):
... |
83b7c6b2d13a748299802172b49f6cd0f51e1e25 | Charnub/python-code | /Python Programming Sheets/Q22.py | 284 | 3.984375 | 4 | myMessage = (input("Enter Message: "))
upperLower = (input("Shout or Whisper?: "))
upperLower = upperLower.lower()
if upperLower == "shout":
myShout = myMessage.upper()
print(myShout)
elif upperLower == "whisper":
myShout2 = myMessage.lower()
print(myShout2)
|
fe3c86ed4282507f3f96c7f070bc1ac0036db547 | Charnub/python-code | /Python Programming Sheets/Q13.py | 301 | 3.796875 | 4 | import random
character = (input("Do you want to create a Character?"))
character = character.lower()
dice1 = random.randint(1,6)
dice2 = random.randint(1,12)
if character == "yes":
calculate = dice2/dice1+10
print("Hit Points: "+(str(calculate)))
elif character == "no":
exit()
|
97ec90e598582256b2ba729af7cde46277481412 | Charnub/python-code | /Test Files/y10help.py | 242 | 3.96875 | 4 | print("Welcome to The Quiz!")
print("Please create a username and password:")
name = (str(input("What is your name?")))
age = (str(input("What is your age?")))
name2 = name[:3]
username = name2+age
print("Your username is: "+username)
|
7134c88c5828fd1675705ccecc1d9d6ad12f446e | Charnub/python-code | /Python Programming Sheets/Q27.py | 145 | 3.609375 | 4 | def coinToss():
import random
if(random.randint(0,1)==0):
return "Heads"
else:
return "Tails"
print(coinToss()) |
90ab247a769e653a54ddc3165106f74e85d83543 | kimstone/itp_week_3 | /day_3/exercise-daniel.py | 597 | 3.828125 | 4 | import requests
import json
# using the requests package, we can make API calls to retrieve JSON
# and storing it into a variable here called "response"
response = requests.get("https://rickandmortyapi.com/api/character")
# verify the response status as 200
# print(response)
# verify the raw string data of the respo... |
1e9988d1becd1cf946dd8d26f614cc0d142edd52 | annilq/python | /chapter-9/admin.py | 482 | 3.625 | 4 | from user import User
class Privileges():
"""docstring for Privileges"""
def __init__(self):
super().__init__()
self.privileges = ["can add post","can delete post","can ban user"]
def show_privileges(self):
print("the admin ",self.privileges)
class Admin(User):
"""docstring for Admin"""
def __init__(self, f... |
f72db317acd78d63bae31468bcece656e696a540 | Nrams1/CSC1015F_Assignment-2 | /pi.py | 332 | 3.984375 | 4 | # calculate pi
# neo ramotlou
# 15 march 2018
x = 0
fraction = 2
import math
p = 2
while x !=2:
x = math.sqrt(x+2)
fraction = (2/x)
p = p*fraction
print('Approximation of pi:', round(p, 3))
radius = eval(input('Enter the radius: \n'))
print('Area:', round((p*(radius**2)), 3))
... |
35cd940e9184686fb80f549dad3555404537714a | daniel321c/coding_quiz | /leafSum.py | 665 | 3.515625 | 4 | # class TreeNode:
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None
class Solution:
def sumNumbers(self, root: TreeNode) -> int:
self.total = 0
self.getNumber(root, 0)
return self.total
def getNumber(self, node, carry):
... |
1ee05d52209e23c2cacd5b512cb3963b26af6b30 | daniel321c/coding_quiz | /filter.py | 710 | 3.625 | 4 | from abc import ABC, abstractmethod
class BaseFilter(ABC):
def __init__(self, field, constraint):
self.field = field
self.constraint = constraint
@abstractmethod
def apply(self, object):
print("i am abstract")
pass
class EquationFilter(BaseFilter):
def __init__(sel... |
076fbda9c9524d527abe372d12f5ab69659bb100 | chz224/Apply-Security-HW1 | /HW_1.py | 760 | 3.90625 | 4 | #Cheng Hao Zheng
#HW 1 Spell Checker
#Github user name: chz224
#import spellchecker library
from spellchecker import SpellChecker
def main():
spell = SpellChecker()
#change the file to the text file you want to spell check
#currently spell checking file MisspelledText.txt
misspelledFile = o... |
f6d7f32228e90e46336cab48bafca14da5be421a | jarbus/epsilon-machine-python | /em/parse_tree_node.py | 4,436 | 3.59375 | 4 | from collections import defaultdict, Counter
from typing import Dict, Sequence, List
from functools import partial
import plantuml
import sys
class ParseTreeNode:
"""Node of a parse tree.
Arguments:
depth: Level of depth in a tree. For roots, depth=0.
target_depth: Denoted as D in Crutchfield... |
4845e44fa0afcea9f4293f45778f6b4ea0da52b0 | jamiegowing/jamiesprojects | /character.py | 402 | 4.28125 | 4 | print("Create your character")
name = input("what is your character's name")
age = int(input("how old is your character"))
strengths = input("what are your character's strengths")
weaknesses = input("what are your character's weaknesses")
print(f"""You'r charicters name is {name}
Your charicter is {age} years old
stren... |
18c4527e5b18282b5d023fe97dbd880bc0f91796 | UPML/python2017 | /62.py | 1,097 | 3.6875 | 4 | import random
def british_word(word, param):
if len(word) >= param + 2:
last_to_change = min(param, len(word) - 2)
positions = random.sample(range(1, len(word) - 1), last_to_change)
characters = []
for j in positions:
characters.append(word[j])
random.shuffle(ch... |
45822bdb6dc53702eff7f2a899117558bb05e4f0 | UPML/python2017 | /4.py | 812 | 3.75 | 4 | def memoize(function):
memo = {}
def wrapper(*args):
if args in memo:
return memo[args]
else:
rv = function(*args)
memo[args] = rv
return rv
return wrapper
# не сложилось=(
@memoize
def fib(n):
if (n == 0):
return 0
if (n ==... |
142ecec208f83818157ce4c8dff7495892e5d5d2 | yagizhan/project-euler | /python/problem_9.py | 450 | 4.15625 | 4 | # A Pythagorean triplet is a set of three natural numbers, a < b < c, for which,
# a2 + b2 = c2
# For example, 32 + 42 = 9 + 16 = 25 = 52.
# There exists exactly one Pythagorean triplet for which a + b + c = 1000.
# Find the product abc.
def abc():
for c in range(1, 1000):
for b in range(1, c):
... |
004571b005999b28b9913db8d646c82f1e562b1b | peterlew/euler-python | /p14.py | 379 | 3.5 | 4 |
results = {1 : 1}
def chainLen(n):
if n in results:
return(results[n])
if n % 2 == 0:
result = 1 + chainLen(n // 2)
else:
result = 1 + chainLen(n * 3 + 1)
results[n] = result
return result
longestChain = 0
longestChainStart = 0
for i in range(1, 1000000):
c = chainLen(i)
if c > longestChain:
longestCh... |
1258ef4008a50145318fbd3e8bd13fabc8989cbb | peterlew/euler-python | /p31.py | 457 | 3.625 | 4 |
coins = [200, 100, 50, 20, 10, 5, 2, 1]
# results dic: (target, max coin index) -> ways to make target
# using coins at max index or greater
results = {}
for i in range(8):
results[(0, i)] = 1
def waysToMake(n, ind):
if (n, ind) in results:
return results[(n, ind)]
total = 0
for i in range(ind, 8):
coin =... |
e7db4be32011e9c773b088453853c24d0c98f4a7 | SoyeonHH/Algorithm_Python | /LeetCode/819.py | 562 | 3.5 | 4 | import collections
import re
paragraph = "Bob hit a ball, the hit BALL flew far after it was hit."
banned = ["hit"]
# Solution
# --------------------------------------------------------------
# 입력값 전처리
words = [word for word in re.sub(r'[^\w]',' ', paragraph)
.lower().split()
if word not in ... |
850cb22e85b943a16b872d0d826b498ab118b90b | SoyeonHH/Algorithm_Python | /LeetCode/206.py | 745 | 3.90625 | 4 | # Definition for singly-linked list.
class ListNode:
def __init__(self, val=0, next=None):
self.val = val
self.next = next
class Solution:
# 재귀 구조로 뒤집기
def reverseList_1(self, head: ListNode) -> ListNode:
def reverse(node: ListNode, prev: ListNode):
if not node... |
77b2c0824d422ba288765da390bd2b7c068bdb62 | Alston-Tang/Spam-Filter | /proc_mail.py | 1,858 | 3.671875 | 4 | # -*- coding: utf-8 -*-
"""
This file contains functions of processing emails
"""
import re
from html2text import html2text
def get_mail_body(_mail_file):
"""
Extract the mail body from the mail file.
:param mail_file: file contains the raw content of email
:type mail_file: str
:return: string th... |
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