动态规划两个数组dp问题系列一＞交错字符串

class Solution {
    public boolean isInterleave(String ss1, String ss2, String ss3) {
        int m = ss1.length(),n = ss2.length(),k = ss3.length();
        ss1 = " "+ ss1; ss2 = " "+ ss2; ss3 = " " + ss3;

        if(m + n != k) return false;

        char[] s1 = ss1.toCharArray(); 
        char[] s2 = ss2.toCharArray(); 
        char[] s3 = ss3.toCharArray();
        //预处理方便对应下标
        boolean[][] dp = new boolean[m+1][n+1];

        //初始化
        dp[0][0] = true;//s1和s2都为空，第一个位置为true
        for(int i = 1; i <= n; i++){
            if(s2[i] == s3[i])
                dp[0][i] = true;
            else break;
        }

        for(int j = 1; j <= m; j++){
            if(s1[j] == s3[j])
                dp[j][0] = true;
            else break;
        }

        for(int i = 1; i <= m; i++)
            for(int j = 1; j <= n; j++){
                // if(s1[i] == s3[i+j] && dp[i-1][j] == true) 
                //     dp[i][j] = dp[i-1][j];
                // else if(s2[j] == s3[i+j] && dp[i][j-1] == true)
                //     dp[i][j] = dp[i][j-1];

                dp[i][j] = (s1[i] == s3[i + j] && dp[i - 1][j]) || (s2[j] == s3[i+j] && dp[i][j - 1]);
            }

        return dp[m][n];    
    }
}