leetcode之Scramble String

原題如下：

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.

這道題目還是有一定難度的，借鑑大牛的博客，下面說一下遞歸求解的思路：

1）如果兩個字符串長度不等，則返回false

2）如果兩個字符串相等，則返回true

3）將兩個字符串排序後比較，如果不相等則返回false

4）將s1分割成左右子樹兩部分，同時將s2分割成左右子樹兩部分，這樣就可以劃分成兩個子問題進行求解，這裏要注意的是還需要交叉驗證，只要有一個滿足就返回true

5）遍歷結束都沒有合適的分割點返回false

class Solution {
public:
    bool isScramble(string s1, string s2) {
		if(s1.size() != s2.size())
			return false;
		if(s1 == s2)
			return true;
		string str1 = s1,str2 = s2;
		sort(str1.begin(),str1.end());
		sort(str2.begin(),str2.end());
		if(str1 != str2)
			return false;
		int len = s1.size();
		for(int i = 1; i < len ; i++){
			string leftS1 = s1.substr(0,i);
			string rightS1 = s1.substr(i);
			string leftS2 = s2.substr(0,i);
			string rightS2 = s2.substr(i);
			if(isScramble(leftS1,leftS2) && isScramble(rightS1,rightS2))
				return true;
			leftS2 = s2.substr(0,len - i);
			rightS2 = s2.substr(len - i);
			if(isScramble(leftS1,rightS2) && isScramble(rightS1,leftS2))
				return true;
		}
		return false;
    }
};