[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.

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