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.

 

有一种方法叫做简单粗暴。。。

 

递归+剪枝

 

bool isScramble(string s1, string s2) {

        // Note: The Solution object is instantiated only once.

    	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'];



		for(int j = 0; j < s2.length(); j++)

			--A[s2[j]-'a'];



		for(int k = 0; k < 26; k++)

			if(A[k] != 0) return false;



        for(int i = 1; i < s1.length(); 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.length()-i, i))

				&& isScramble(s1.substr(i), s2.substr(0,s2.length()-i)));

			if(result) return true;

		}

		return false;

    }