URL:https://leetcode.com/problems/regular-expression-matching 解法 动态规划。 p[i - 1] == s[i - 1], dp[i][j] = dp[i - 1][j - 1]:字符匹配,如 'a' 匹配 'a',两个字符匹配的情况下,看两个字符之前字符的匹配情况。 p[i - 1] = '.', dp[i][j] = dp[i - 1][j - 1]...
【LeetCode】10. Regular Expression Matching 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). The func...
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 ...
'.' Matches any single character. '*' Matches zero or more of the preceding element. The matching should cover the entire 1. 2. 3. 4. 方法:递归 递归逐个字符比较:每次要比较的时候先判断比较的字符后面的一个字符是不是'*', 例如A=aa和B=a**b是否匹配的时候,在判断第一个字符是否匹配之前先...
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: ...
10. 正则表达式匹配 - 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"
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. ...
Recursion has helped code become simple. 首先说下我犯下的错误吧。 最严重的错误,给出的两个字符串,s , p, 其实只会在p中出现 "." and "". 因为是表达式匹配,所以s中不会出现这些符号。 然后就是对 ''含义的理解。"*"表示的是,前面那个字符出现的次数。
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: ...