leetcode-regular-expression-match

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题目大意

　　https://leetcode.com/problems/regular-expression-matching/

　　跟这个题类似https://leetcode.com/problems/wildcard-matching/，只是匹配规则变化

题目分析

　　记忆化搜索，详细解释可以看注释

代码

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public class Solution {
    
    private char[][] dp;
    
    private char search(String s, int si, String p, int pi) {
        
        int lenS = s.length();
        int lenP = p.length();
        
        if (si == lenS && pi == lenP) {
            return dp[si][pi] = 1;
        }
        
        if (dp[si][pi] > 0) {
            return dp[si][pi];
        }
        
        char a = 0;
        if (si == lenS) {
            if (p.charAt(pi) == '*') {
                return dp[si][pi] = search(s, si, p, pi + 2);
            } else {
                // 一定匹配不上
                return dp[si][pi] = 2;
            }
        } else {
            a = s.charAt(si);
        }
        
        char b = 0;
        if (pi == lenP) {
            // 一定匹配不上
            return dp[si][pi] = 2;
        } else {
            b = p.charAt(pi);
        }
        
        if (b != '*' && b != '.') { // b不是*或者i
            if (a != b) {
                return dp[si][pi] = 2;
            }
            return dp[si][pi] = search(s, si + 1, p, pi + 1);
        }
        if (b == '*') {
            if (search(s, si, p, pi + 2) == 1) { //出现星号，先尝试匹配空
                return dp[si][pi] = 1;
            }
            for (int loc = si; loc < lenS; loc++) { //再尝试匹配1-n个
                if (s.charAt(loc) != p.charAt(pi + 1) && p.charAt(pi + 1) != '.') { // 注意.*情况
                    break;
                }
                if (search(s, loc + 1, p, pi + 2) == 1) {
                    return dp[si][pi] = 1;
                }
            }
            return dp[si][pi] = 2;
        }
        if (b == '.') {
            return dp[si][pi] = search(s, si + 1, p, pi + 1);
        }
        return dp[si][pi] = 2;
    }
    
    public boolean isMatch(String s, String p) {
        if (s == null || p == null) {
            return false;
        }
        s = new StringBuilder(s).reverse().toString();
        p = new StringBuilder(p).reverse().toString();
        int lenS = s.length();
        int lenP = p.length();
        dp = new char[lenS + 1][lenP + 1];
        for (int i = 0; i <= lenS; i++) {
            for (int j = 0; j <= lenP; j++) {
                dp[i][j] = 0;
            }
        }
        return search(s, 0, p, 0) == 1 ? true : false;
    }
}