Leetcode - Scramble String

Given a string s1, we may represent it as a binary tree by partitioning it to two non-empty substrings recursively.

Below is one possible representation of s1 = "great":

    great
   /    \
  gr    eat
/ \    /  \
g   r  e   at
           / \
          a   t

To scramble the string, we may choose any non-leaf node and swap its two children.

For example, if we choose the node "gr" and swap its two children, it produces a scrambled string "rgeat".

    rgeat
   /    \
  rg    eat
/ \    /  \
r   g  e   at
           / \
          a   t

We say that "rgeat" is a scrambled string of "great".

Similarly, if we continue to swap the children of nodes "eat" and "at", it produces a scrambled string "rgtae".

    rgtae
   /    \
  rg    tae
/ \    /  \
r   g  ta  e
       / \
      t   a

We say that "rgtae" is a scrambled string of "great".

Given two strings s1 and s2 of the same length, determine if s2 is a scrambled string of s1.

[balabala] 暴力解决枚举出s1的全部scramble string, 存在大量重复计算；分段比较s1,s2也存在大量重复计算，比如s1=abcdef, s2=dcabef, isScramble(bcdef, cabef) 就会重复计算isScramble(cdef, abef)。使用动态规划，保存s1, s2每个位置处各种长度对应的字串的匹配情况，dp[l][i][j]表示s1[i, i+l) 和 s2[j, j+l)是否是scramble关系。

// timeout
public boolean isScramble1(String s1, String s2) {
        if (s1 == null || s2 == null || s1.length() != s2.length())
            return false;
        if (s1.equals(s2))
            return true;
        int lengthS1 = s1.length();
        for (int i = 1; i < lengthS1; i++) {
            boolean cond1 = isScramble(s1.substring(0, i), s2.substring(0, i)) 
                    && isScramble(s1.substring(i, lengthS1), s2.substring(i, lengthS1));
            boolean cond2 = isScramble(s1.substring(0, i), s2.substring(lengthS1 - i, lengthS1)) 
                    && isScramble(s1.substring(i, lengthS1), s2.substring(0, lengthS1 - i));
            if (cond1 || cond2)
                return true;
        }
        return false;
    }
    
    public boolean isScramble(String s1, String s2) {
        if (s1 == null || s2 == null || s1.length() != s2.length())
            return false;
        if (s1.equals(s2))
            return true;
        int length = s1.length();
        boolean[][][] dp = new boolean[length + 1][length][length];
        for (int l = 1; l <= length; l++) {
            for (int i = 0; i <= length - l; i++) {
                for (int j = 0; j <= length - l; j++) {
                    if (l == 1) {
                        dp[1][i][j] = s1.charAt(i) == s2.charAt(j);
                    } else {
                        for (int k = 1; k < l; k++) {
                            dp[l][i][j] = (dp[k][i][j] && dp[l - k][i + k][j + k])
                                            || (dp[k][i][j + l - k] && dp[l - k][i + k][j]);
                            if (dp[l][i][j])
                                break;
                        }
                    }
                }
            }
        }
        return dp[length][0][0];
}