Leetcode(10)正则表达式匹配 [题目表述]: 给定一个字符串 (s) 和一个字符模式 (p).实现支持 '.' 和 '*' 的正则表达式匹配. '.' 匹配任意单个字符. '*' 匹配零个或多个前面的元素. 匹配应该覆盖整个字符串 (s) ,而不是部分字符串. 第一次:未完成(未能解决.*问题 class Solution: def isMatch(self, s: str, p: str) -> bool: index_s=0 index_p=0 form_num='' if len…

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…

Hard! 题目描述: 给定一个字符串 (s) 和一个字符模式 (p).实现支持 '.' 和 '*' 的正则表达式匹配. '.' 匹配任意单个字符. '*' 匹配零个或多个前面的元素. 匹配应该覆盖整个字符串 (s) ,而不是部分字符串. 说明: s 可能为空,且只包含从 a-z 的小写字母. p 可能为空,且只包含从 a-z 的小写字母,以及字符 . 和 *. 示例 1: 输入: s = "aa" p = "a" 输出: false 解释: "a&quo…

[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] &&…

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…

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…

问题描述: Given two arrays, write a function to compute their intersection. Example:Given nums1 = [1, 2, 2, 1], nums2 = [2, 2], return [2]. Note: Each element in the result must be unique. The result can be in any order. my answer: 思路: 遍历nums1里的元素,看nums2…

题目:匹配正则表达式 题目难度:hard 题目内容: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 fu…

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 function prototype should…