10-week-algorithm-excercise-master/Week_09/README.md

学习笔记

#	Title	Solutions
32	longest-valid-parentheses	动态规划(Go,Py),栈(Go,Py),贪心(Go,Py)
64	minimum-path-sum	动态规划(Go,Py)
72	edit-distance	动态规划(Go,Py)
85	maximal-rectangle	动态规划(Go,Py)
115	distinct-subsequences	动态规划(Go,Py)
58	length-of-last-word	逆序迭代(Go,Py)
709	to-lower-case	迭代(Go,Py)
8	string-to-integer-atoi	迭代(Go,Py)
14	longest-common-prefix	迭代(Go,Py)
344	reverse-string	双指针(Go,Py)
541	reverse-string-ii	迭代(Go,Py)
557	reverse-words-in-a-string-iii	迭代(Go,Py)
917	reverse-only-letters	双指针(Go,Py)
438	find-all-anagrams-in-a-string	暴力(Go,滑动窗口(Go,Py)
125	valid-palindrome	双指针(Go,Py)
680	valid-palindrome-ii	双指针夹逼(Go,Py)
5	longest-palindromic-substring	动态规划(Go,Py),中心扩展(Go,Py)

题解

32. longest-valid-parentheses

1. 动态规划：
   • 记录s[i] == ')'的dp[i]
     • 当s[i-1] == '(': dp[i] = dp[i-2] + 2
     • 当s[i-1] == '('且dp[i-1] > 0: dp[i] = dp[i-1] + dp[i-dp[i-1]-2] + 2
2. 栈:
   • 用栈记录下标，初始加一个-1的值[-1]
   • 当s[i] == '('时，i入栈， 否则弹出栈顶元素，之后当前栈顶即为左边界:len = i - stack[len(stack)-1] 3: 贪心:
   • 用left和right分别记录左右括号的数量
   • 先从左向右遍历，遇到'('则left++,否则right++,当left==right为可能的解，right>left时将left/right重置为0
   • 从右向左遍历亦然

72. edit-distance

1. 动态规划：
   • 当word1[i-1] == word2[j-1]时 dp[i][j] = dp[i-1][j-1],否则 min(dp[i - 1][j - 1], dp[i][j - 1], dp[i - 1][j]) + 1

85. maximal-rectangle

1. 动态规划：
   • 使用三个数组left[n], right[n],height[n]分别记录当前的左右边界和高

115. distinct-subsequences

1. 动态规划：
   • 当s[i-1] != t[j-1]时： dp[i][j] = dp[i-1][j] (不取s[j]) 否则 dp[i][j] = dp[i-1][j] + dp[i-1][j-1]