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]...
保证每次出现字符*时,前面都匹配到有效的字符 题目链接:https://leetcode.cn/problems/regular-expression-matching/description/ 『1』动态规划 解题思路: 参考题解:『 动态规划 』字符串 DP 详解:状态转移推导 + 滚动优化 + 提前结束 实现代码: classSolution{ // Dynamic Programming // N is the length of ...
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: bool isMatch(const char *s, c...
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: ...
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: ...
Regular Expression Matching@LeetCode Regular Expression Matching 比较典型的动态规划题,重点就在于*号的匹配上。 分三步走: 当模式串为空时。检查内容串是否为空(为空则返回true,反之返回false)。注意这里反过来是不成立的,也就是不能检查当内容穿为空时,模式串是否为空,因为如果模式串最后一个字符是*,那么...
原题链接:http://oj.leetcode.com/problems/regular-expression-matching/ 这个题目比较常见,但是难度还是比较大的。我们先来看看brute force怎么解决。基本思路就是先看字符串s和p的从i和j开始的子串是否匹配,用递归的方法直到串的最后,最后回溯回来得到结果。假设现在走到s的i位置,p的j位置,情况分为下列两种: ...