[leetcode]Scramble String

Scramble String

题目：

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.

这道题，题目都没读懂是什么意思~~，为自己的理解能力捉急~~，参考了http://blog.unieagle.net/2012/10/23/leetcode%E9%A2%98%E7%9B%AE%EF%BC%9Ascramble-string%EF%BC%8C%E4%B8%89%E7%BB%B4%E5%8A%A8%E6%80%81%E8%A7%84%E5%88%92/和http://blog.theliuy.com/scramble-string/这两篇博客，看了代码才懂题的意思。用的就是动态规划。

c++代码：

#include <string>
#include <iostream>
#include <vector>
using namespace std;
class Solution {
public:
    int char_size;

    bool isScramble(string s1, string s2) {
		char_size = 26;
        return isScrambleHelper(s1, s2);
    }
    
    bool isScrambleHelper(string &s1, string &s2) {
        if (s1.size() != s2.size())
            return false;
        if (s1 == s2)
            return true;
        
        int size = s1.size();
        vector<int> bucket(char_size, 0);
        string s11, s12, s21, s22;
        
        // Check wheter they have the same chars
        for (int i = 0; i < s1.size(); ++i) {
            bucket[s1[i] - 'a'] += 1;
            bucket[s2[i] - 'a'] -= 1;
        }
        for (int i = 0; i < char_size; ++i) {
            if (bucket[i] != 0)
                return false;
        }
        

        for (int i = 1; i < size; ++i) {
            s11 = s1.substr(0, i);
            s12 = s1.substr(i);
            
            s21 = s2.substr(0, i);
            s22 = s2.substr(i);
            if (isScrambleHelper(s11, s21) && isScrambleHelper(s12, s22))
                return true;
                
            s21 = s2.substr(size - i);
            s22 = s2.substr(0, size - i);
            if (isScrambleHelper(s11, s21) && isScrambleHelper(s12, s22))
                return true;
        }
        
        return false;
    }

};
int main()
{
	Solution s = Solution();
	bool r = s.isScramble("great", "rgeat");
	cout << r << endl;
}