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.

只想到了递归，但总是围绕着树来思考，不得其所。google之后借鉴了同学们的，放在这里，只是为了以后自己回顾。

#include <algorithm>


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

==================第二次=================

注意剪枝啊，一味的递归会超时。

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