44. Wildcard Matching 

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class Solution {
    public boolean isMatch(String s, String p) {
        int m = s.length(), n = p.length();

        // 定义 dp[i][j] 为 s[0, i - 1] 与 p[0, j - 1] 是否能够匹配
        boolean[][] dp = new boolean[m + 1][n + 1];
        dp[0][0] = true;

        for (int j = 1; j <= n && p.charAt(j - 1) == '*'; j++) {
            dp[0][j] = true;
        }

        for (int i = 1; i <= m; i++) {
            for (int j = 1; j <= n; j++) {
                char c = p.charAt(j - 1);
                if (c == '*') {
                    // 匹配一个空字符串
                    dp[i][j] = dp[i][j - 1];
                    // 匹配一个字符
                    dp[i][j] |= dp[i - 1][j - 1];
                    // 匹配多个字符，不要忘记此场景
                    dp[i][j] |= dp[i - 1][j];
                } else if (c == '?') {
                    dp[i][j] = dp[i - 1][j - 1];
                } else {
                    dp[i][j] = c == s.charAt(i - 1) && dp[i - 1][j - 1];
                }
            }
        }

        return dp[m][n];
    }
}

References

44. Wildcard Matching