题解 | #通配符匹配#

    public boolean isMatch(String s, String p) {
        if( p == null || p.length() == 0){
            return s == null || s.length() == 0;
        }
        int sLen = s.length();
        int pLen = p.length();
        //dp[i][j]  s的 0... i-1 元素 能够被 p的 0....j-1 元素匹配
        boolean[][] dp = new boolean[sLen + 1][pLen + 1];
        dp[0][0] = true;
        
        for(int j = 1;j<=pLen;j++){
            dp[0][j] = p.charAt(j-1) == '*' && dp[0][j-1];
        }
        for(int i = 1;i<= sLen;i++){
            for(int j = 1;j <=pLen;j ++){
                char a = s.charAt(i-1);
                char b = p.charAt(j-1);
                if(a == b || b == '?'){
                    dp[i][j] = dp[i-1][j-1];
                }else if(b == '*'){
                    dp[i][j] = dp[i-1][j] || dp[i][j-1];
                }
            }
        }
        return dp[sLen][pLen];
    }
}