LEETCODE: Scramble String

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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.

大招就是递归啊，换来换去其实还有有些不变的。

详细参考：http://www.blogjava.net/sandy/archive/2013/05/22/399605.html

class Solution {
public:
    bool isScramble(string s1, string s2) {
        if(s1.length() != s2.length()) return false;
        if(s1 == s2) return true;
        string s1copy(s1);
        string s2copy(s2);
        sort(s1copy.begin(), s1copy.end());
        sort(s2copy.begin(), s2copy.end());
        if(s1copy != s2copy) return false;
        
        for(int ii = 1; ii < s1.length(); ii ++) {
            bool result = isScramble(s1.substr(0, ii), s2.substr(0, ii)) && isScramble(s1.substr(ii), s2.substr(ii));
            result = result || (isScramble(s1.substr(0, ii), s2.substr(s2.length() - ii, ii)) && isScramble(s1.substr(ii), s2.substr(0, s2.length() - ii)));
            if(result) return true;
        }
        
        return false;
    }
};