[LeetCode LintCode] 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.

Recursive:

class Solution {
public:
    bool isScramble(string s1, string s2) {
        int l1 = s1.length();
        int l2 = s2.length();
        if(l1 != l2) return false;
        
        if(l1==1) return s1[0] == s2[0];
        
        int A[26] = {0};
        for(int i=0;i<l1;i++)
        {
            A[s1[i]-'a']++;
            A[s2[i]-'a']--;
        }
        for(int i=0;i<26;i++)
        {
            if(A[i]!=0) return false;
        }
        for(int i=1;i<l1;i++)
        {
            bool res = (isScramble(s1.substr(0,i), s2.substr(0,i)) && 
                            isScramble(s1.substr(i), s2.substr(i))) ||
                       (isScramble(s1.substr(0,i), s2.substr(l1-i)) &&
                            isScramble(s1.substr(i), s2.substr(0,l1-i)));
            if(res) return true;
        }
        return false;
    }
};