Given a strings1, we may represent it as a binary tree by partitioning it to two non-empty substrings recursively.

Below is one possible representation ofs1="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 stringss1ands2of the same length, determine ifs2is a scrambled string ofs1.

tag: DFS, divide and conquer, recursion

挺好的divide and conquer

class Solution {
    public boolean isScramble(String s1, String s2) {
        if (s1.equals(s2)) return true;
        if (s1.length() != s2.length()) return false;

        int[] hash = new int[26];

        for (int i = 0; i < s1.length(); i++){
            hash[s1.charAt(i) - 'a']++;
            hash[s2.charAt(i) - 'a']--;
        }

        for (int i = 0; i < 26; i++){
            if (hash[i] != 0) return false;
        }

        for (int i = 1; i < s1.length(); i++){
            if (isScramble(s1.substring(0, i), s2.substring(0, i)) && isScramble(s1.substring(i), s2.substring(i))) return true;
            if (isScramble(s1.substring(0, i), s2.substring(s2.length() - i)) && isScramble(s1.substring(i), s2.substring(0, s2.length() - i))) return true;
        }

        return false;
    }
}