leetcode44

class Solution {
    public boolean isMatch(String s, String p) {
        if (p == null || p.length() == 0) {
            return s == null || s.length() == 0;
        }
        boolean[][] dp = new boolean[s.length() + 1][p.length() + 1];
        dp[0][0] = true;
        int i, j;

        for (i = 1; i <= p.length(); i++) {
            if (p.charAt(i - 1) == '*') {
                dp[0][i] = true;
            } else {
                break;
            }
        }

        for (i = 1; i <= s.length(); i++) {
            for (j = 1; j <= p.length(); j++) {
                // the letter of p[j] exactly match with s[i] or is '?'
                if (s.charAt(i - 1) == p.charAt(j - 1) || p.charAt(j - 1) == '?') {
                    dp[i][j] = dp[i - 1][j - 1];
                }

                // the letter of p is '*'
                if (p.charAt(j - 1) == '*') {
                    /*
                     * dp[i][j] = dp[i][j-1] means '*' act as empty card dp[i][j] = dp[i-1][j] means
                     * '*' act as wild card to match with s[i]
                     */
                    dp[i][j] = dp[i][j - 1] || dp[i - 1][j];
                }
            }
        }
        return dp[s.length()][p.length()];
    }
}

参考：https://leetcode.com/problems/wildcard-matching/discuss/429337/Java%3A-Wild-card!

补充一个python的实现：

class Solution:
    def isMatch(self, s: str, p: str) -> bool:
        m,n = len(s),len(p)
        dp = [[False for _ in range(n+1)]for _ in range(m+1)]
        dp[0][0] = True
        
        for j in range(1,n+1):
            if p[j-1] == '*':
                dp[0][j] = True
            else:
                break
        for i in range(1,m+1):
            for j in range(1,n+1):
                if s[i-1] == p[j-1] or p[j-1] == '?':
                    dp[i][j] = dp[i-1][j-1]
                elif p[j-1] == '*':
                    dp[i][j] = dp[i-1][j] or dp[i][j-1]
        # print(dp)
        return dp[m][n]

思路：动态规划。

举例分析：s='acdcb' p='a*c?b' 。s长度为m，p长度为n。

初始化二维数组dp，有m+1行，n+1列。

dp[0][0] = True，先设置第0行，如果p中对应字符是‘*’，则设置为True，遇到非'*'后面都设置为False。

从第1行开始，如果s与p中对应字符是相同或者p中是'?'，则dp取其'左上角'元素的值。

如果p中的字符是'*'，则dp判断其'上方'和'左侧'的逻辑或，(即上方或左侧是否为True)。