【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 precedin
(Version 1.1) (Regular Expression Matching在LeetCode上的标签是DP,但其实不用DP只用recursion也可以过,这题的test case一般。) 这一题自己没有做出来,感觉还是思维角度的问题。个人感觉一个可行的思维角度是先从Code Ganker的那个recursive的角度出发(http://blog.csdn.net/linhuanmars/article/details/21145563),...
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]=...
leetcode.10---Regular Expression Matching '.' and '*'. '.' 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...
public class RegularExpressionMatching { public static void main(String[] args) { System.out.println(isMatch("aa","a")); System.out.println(isMatch("aa","aa")); System.out.println(isMatch("aaa","aa")); System.out.println(isMatch("aa", "a*")); ...
Loading...leetcode.com/problems/regular-expression-matching/discuss/5684/C%2B%2B-O(n)-space-DP 打个比方,从A走到B,第一种写法是一直走下去,直到确定当前走法行不通,返回上次可选择的地方换一种走法,如果所有走法皆不通,则退出本层循环,回到上上次可选择的地方但假如走通了,就直接返回答案。但第...
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. ...
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 inpu ...
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