44. Wildcard Matching

最新推荐文章于 2020-10-29 13:48:23 发布

jimhust

最新推荐文章于 2020-10-29 13:48:23 发布

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分类专栏： leetcode 文章标签： leetcode

本文链接：https://blog.youkuaiyun.com/jimhust/article/details/58587369

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本文介绍了一种解决正则表达式匹配问题的算法，该算法使用动态规划的方法来判断输入字符串是否能被给定的包含'?'和'*'的模式完全匹配。文章详细解释了动态规划状态转移方程，并提供了完整的Java实现代码。

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'?' Matches any single character.
'*' Matches any sequence of characters (including the empty sequence).

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", "*") → true
isMatch("aa", "a*") → true
isMatch("ab", "?*") → true
isMatch("aab", "c*a*b") → false

we use dp to solve this problem;

let dp[i][j] be the boolean that means whether S(0 -> i) and P(0 -> j) is a match:

if(P[j] == '?') dp[i][j] = dp[i - 1][j - 1];

if(P[j] == '*') dp[i][j] = dp[i - 1][j] || dp[i][j - 1];

else dp[i][j] = (S[i] == P[j]) && dp[i - 1][j - 1];

Initialization: dp[0][0] = true;

dp[i][0] = false;

dp[0][j] 如果j之前都是＊就为true

code:

public class Solution {
    public boolean isMatch(String s, String p) {
        boolean[][] dp = new boolean[s.length() + 1][p.length() + 1];
        dp[0][0] = true;
        for(int i = 0; i < p.length(); i++){
            if(p.charAt(i) == '*') dp[0][i + 1] = true; 
            else break;
        }
        
        
        for(int i = 1; i<=s.length(); i++){
            for(int j = 1; j<=p.length(); j++){
                char c = p.charAt(j - 1);
                if(c == '?'){
                    dp[i][j] = dp[i - 1][j - 1];
                } else if (c == '*'){
                    dp[i][j] = dp[i - 1][j] || dp[i][j - 1];
                } else {
                    dp[i][j] = (s.charAt(i - 1) == c) && dp[i - 1][j - 1];
                }
            }
        }
        
        return dp[s.length()][p.length()];
    }
}