[LeetCode #10] Regular Expression Matching

最新推荐文章于 2022-05-01 16:20:00 发布

转载最新推荐文章于 2022-05-01 16:20:00 发布 · 55 阅读

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原文链接：http://www.cnblogs.com/amadis/p/5926445.html

本文介绍了一种实现正则表达式匹配的算法，支持 '.' 和 '*' 的使用。该算法通过动态规划解决字符串完全匹配的问题，并提供了详细的 C++ 实现代码及示例。

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, const char *p)

Some examples:
isMatch("aa","a") → false
isMatch("aa","aa") → true
isMatch("aaa","aa") → false
isMatch("aa", "a*") → true
isMatch("aa", ".*") → true
isMatch("ab", ".*") → true
isMatch("aab", "c*a*b") → true

 1 class Solution {
 2 public:
 3     bool isMatch(string s, string p) {
 4         int m = s.size(), n = p.size();
 5         
 6         vector<vector<bool>> dp (m+1, vector<bool> (n+1, false));
 7         
 8         dp[0][0] = true;
 9         
10         for (int i = 0; i < m; i++){
11             dp[i+1][0] = false;
12         }
13         
14         for (int j = 0; j < n; j++){
15             dp[0][j+1] = j > 0 && p[j] == '*' && dp[0][j-1];
16         }
17         
18         for (int i = 0; i < m; i++){
19             for (int j = 0; j < n; j++){
20                 if (p[j] != '*'){
21                     dp[i+1][j+1] = dp[i][j] && (s[i] == p[j] || p[j] == '.');
22                 }else{
23                     dp[i+1][j+1] = dp[i+1][j-1] || ((dp[i][j+1]) && (s[i] == p[j-1] || p[j-1] == '.'));
24                 }
25             }
26         }
27         
28         return dp[m][n];
29     }
30 };