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. 

分析：这道题用递归的方法解逻辑清晰且不易出错，我们将原来的字符串s1分为左边长为i，右边长为l-i (l为字符串总长度)两部分，用递归的思想，如果下面两种情况一种成立则s2是s1的scramble string。

1）s1左端长为i的子串与s2左端长为i的子串互为scramble string， s1右端长为l-i的子串与s2右端长为l-i的子串互为scramble string。

2）s1左端长为i的子串与s2右端长为i的子串互为scramble string, s1右端长为l-i的子串与s2左端长为l-i的子串互为scramble string。

代码如下：

class Solution {
public:
    bool isScramble(string s1, string s2) {
        if(s1 == s2) return true;
        //if characters in s1 and s2 are different, return false
        string str1 = s1, str2 = s2;
        sort(str1.begin(), str1.end());
        sort(str2.begin(), str2.end());
        if(str1 != str2) return false;
        int l = s1.length();
        //iterate length of left-edian substring of s1
        for(int i = 1; i <= l - 1; i++){
            if(isScramble(s1.substr(0,i), s2.substr(l - i))&&isScramble(s1.substr(i), s2.substr(0, l-i)) || isScramble(s1.substr(0,i), s2.substr(0,i))&&isScramble(s1.substr(i), s2.substr(i))) return true;
        }
        
        return false;
    }
};