# -*- coding: utf8 -*-'''https://oj.leetcode.com/problems/regular-expression-matching/ Implement regular expression matching with support for '.' and '*'. '.' Matches any single character.'*' Matches zero or more of the preceding element. The matchin…

我现在在做一个叫<leetbook>的免费开源书项目,力求提供最易懂的中文思路,目前把解题思路都同步更新到gitbook上了,需要的同学可以去看看 书的地址:https://hk029.gitbooks.io/leetbook/ 010. Regular Expression Matching 问题 Implement regular expression matching with support for ‘.’ and ‘*’. ‘.’ Matches any single charact…

Implement regular expression matching with support for '.' and '*'. '.' Matches any single character. '*' Matches zero or more of the preceding element. The matching should cover the entire input string (not partial). The function prototype should be…

一天一道LeetCode系列 (一)题目 Implement regular expression matching with support for '.' and '*'. '.' Matches any single character. '*' Matches zero or more of the preceding element. The matching should cover the entire input string (not partial). The functio…

Given an input string (s) and a pattern (p), implement regular expression matching with support for '.' and '*'. '.' Matches any single character. '*' Matches zero or more of the preceding element. The matching should cover the entire input string (n…

Implement regular expression matching with support for '.' and '*'. '.' Matches any single character. '*' Matches zero or more of the preceding element. The matching should cover the entire input string (not partial). The function prototype should be…

题目描述: Implement regular expression matching with support for '.' and '*'. '.' Matches any single character. '*' Matches zero or more of the preceding element. 解题思路: 这道题如果只考虑“.”的话其实很好完成,所以解题的关键在于处理“*”的情况.以为“*”与前一个字母有关,所以应该整体考虑ch*……的情况.ch*可以匹配0-n个s的字符串…

Implement regular expression matching with support for '.' and '*'. '.' Matches any single character. '*' Matches zero or more of the preceding element. The matching should cover the entire input string (not partial). The function prototype should be…

[题意] 给两个字符串s和p,判断s是否能用p进行匹配. [题解] dp[i][j]表示s的前i个是否能被p的前j个匹配. 首先可以分成3大类情况,我们先从简单的看起: (1)s[i - 1] = p[j - 1],易得dp[i][j] = dp[i-1][j-1] (2)p[i - 1] = '.',因为'.'可以匹配任何字符,所以dp[i][j] = dp[i-1][j-1] (3)p[i - 1] = '*',这种情况就比较复杂了. 当p[j - 2] != s[i - 1] &&…

原题地址:https://oj.leetcode.com/problems/regular-expression-matching/ 题意: Implement regular expression matching with support for '.' and '*'. '.' Matches any single character. '*' Matches zero or more of the preceding element. The matching should cover…