Regular Expression Matching - LeetCode 注意点 “.*”可以匹配任何字符串(包括空的) 解法 解法一:参考Regular Expression Matching 正则表达式匹配。时间复杂度为O(n) classSolution{public: bool firstCharIsMatch(char s,char p) {if(s == p || p =='.') {returntrue; }returnfalse; } bool isMatch(...
代码: classSolution {publicbooleanisMatch(String s, String p) {if(s==null|| p==null)returnfalse;boolean[][] dp=newboolean[s.length()+1][p.length()+1]; dp[0][0]=true;for(inti=0;i<p.length();i++){if(p.charAt(i)=='*' && dp[0][i-1]==true){// 当i=0时,p.charAt...
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]=...
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 (not partial).Note: s could be empty ...
leetcode 10. Regular Expression Matching 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)....
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); ...
给你一个字符串s和一个字符规律p,请你来实现一个支持'.'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那一个元素 所谓匹配,是要涵盖整个字符串s的,而不是部分字符串。 示例1: 输入:s = "aa", p = "a"输出:false解释:"a" 无法匹配 "aa" 整个字符串。
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. ...
s = "ab" p = ".*" Output: true Input: s = "aab" p = "c*a*b" Output: true Input: s = "mississippi" p = "mis*is*p*." Output: false Solution 直观的可以采用递归来做: 如果p第二个字符是*,表示可以匹配0个或者多个: 从匹配0个开始尝试,即匹配s与p[2:](刚开始写一直出错,因为没...
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