Implement regular expression matching with support for'.'and'*'. '.' Matches any single character. '*' Matches zero or more of the preceding element. The matching should cover theentireinput string (not partial)
乘风破浪:LeetCode真题_010_Regular Expression Matching 一、前言 关于正则表达式我们使用得非常多,但是如果让我们自己写一个,却是有非常大的困难的,我们可能想到状态机,确定,非确定状态机确实是一种解决方法,不过需要消耗很大的时间去推理和计算,对于正则表达式的
10. Regular Expression Matching #1 动态规划[AC] 在正则表达式中,由于不同的字符会有不同的匹配规则,其中.和*比较特别,需要对这两类字符进行分类讨论。 定义状态dp[i][j]表示输入串长度为i,模式串长度为j时,是否能匹配。 初始化状态值: 输入串为空,模式串为空: dp[0][0]必然可以匹配,即dp[0][0]=...
10. Regular Expression Matching 难度:hard Share 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 theentireinput strin...
classSolution{ public: boolisMatch(constchar*s,constchar*p) { if(*p=='\0')return*s=='\0'; //next char is not '*':must match current character if(*(p+1)!='*') { if(*p==*s||(*p=='.'&&*s!='\0')) returnisMatch(s+1,p+1); ...
'.': 匹配任何单字符 '*': 匹配0个或者多个前置元素(字符串匹配) Java Solution 题目链接 https://leetcode.com/problems/regular-expression-matching/?tab=Description '.' Matches any single character.匹配任何单字符 '*' Matches zero or more of the preceding element.匹配0个或者多个前置元素 ...
Regular Expression Matching 解法一:简单暴力求解 解题思路 对模板字符串进行处理,分为四种情况: 1)字母* 2).* 3)字母 4). 深度遍历求解 代码 class Solution{public:typedefintMType;enum Roster{LetterAsterisk=0,DotAsterisk,Letter,Dot};intmodel[101][2];inttrans(stringp){intlen=p.length();intindex=...
地址:https://github.com/hk029/leetcode 这个是书的地址:https://hk029.gitbooks.io/leetbook/ 这里写图片描述 Regular Expression Matching 问题 Implement regular expression matching with support for '.' and '*'. '.' Matches any single character. ...
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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: ...