标签：leetcode   算法   

问题描述：

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.

基本思想：

递归的分别验证树的左子树和右子树。

代码：

   public boolean isScramble(String s1, String s2) {  //java
        if(s1 == null || s2 == null || s1.length() != s2.length())
            return false;
        
        if(s1.equals(s2))
            return true;
        char [] chs1= s1.toCharArray();
        char [] chs2= s2.toCharArray();
        Arrays.sort(chs1);
        Arrays.sort(chs2);
        if(!Arrays.equals(chs1, chs2))
        	return false;
        	
        for(int i = 1; i < s1.length(); i++)
        {
            if(isScramble(s1.substring(0,i),s2.substring(0,i))&&isScramble(s1.substring(i),s2.substring(i))) 
                return true;
            if(isScramble(s1.substring(0,i),s2.substring(s2.length()-i))&&isScramble(s1.substring(i),s2.substring(0,s2.length()-i)))
                return true;
        }
        return false;
    }

[leetcode]Scramble String

标签：leetcode   算法   

原文地址：http://blog.csdn.net/chenlei0630/article/details/42434645