[Leetcode] 87. 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.

public class Solution {    public boolean isScramble(String s1, String s2) {        if(s1 == null || s2 == null || s1.length() != s2.length()) return false;        boolean[][][] res = new boolean[s1.length()][s2.length()][s1.length() + 1];        for(int i = 0; i < s1.length(); i++){            for(int j = 0; j < s2.length(); j++){                  res[i][j][1] = s1.charAt(i) == s2.charAt(j);              }          }          for(int len = 2; len <= s1.length(); len++){              for(int i = 0; i < s1.length() - len + 1; i++){                for(int j = 0; j < s2.length() - len + 1; j++){                    for(int k = 1; k < len; k++){                        res[i][j][len] |= (res[i][j][k] && res[i + k][j + k][len - k])                                        || (res[i][j + len - k][k] && res[i + k][j][len - k]);                      }                }              }          }          return res[0][0][s1.length()];    }}