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(std::string s1, std::string s2) {        if(s1.size() != s2.size()) return false;if(s1 == s2) return true;int vec[26] = {0};for (int i = 0; i < s1.size(); i++){++vec[s1[i] - 'a'];--vec[s2[i] - 'a'];}for (int i = 0; i < 26; i++){if(vec[i]) 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;    }};