[leecode]Scramble String

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.

算法思路：

分治法：其实就是暴力法= =///

代码如下：

public class Solution {
    public boolean isScramble(String s1,String s2){
        if(s1.length() != s2.length()) return false;
        if(s1.length() <= 1 && s1.equals(s2)) return true;
        if(s1.length() == 2){
            if(s1.equals(s2) ) return true;
            if(s1.charAt(0) == s2.charAt(1) && s1.charAt(1) == s2.charAt(0)) return true;
            return false;
        }
        char[] s1Array = s1.toCharArray();
        char[] s2Array = s2.toCharArray();
        Arrays.sort(s2Array);
        Arrays.sort(s1Array);
        for(int i = 0; i < s1.length(); i++){
            if(s1Array[i] != s2Array[i])
                return false;
        }
        boolean result = false;
        for(int i = 1; i < s1.length();i++){
            String s1pre = s1.substring(0,i);
            String s1suf = s1.substring(i);
            String s2pre = s2.substring(0, i);
            String s2suf = s2.substring(i);
            result = isScramble(s1pre, s2pre) && isScramble(s1suf, s2suf);
            if(!result){
                String s3pre = s2.substring(0, s1.length() - i);
                String s3suf = s2.substring(s1.length() - i);
                result = isScramble(s1pre, s3suf) && isScramble(s1suf, s3pre);
            }
            if(result) return true;
        }
        return result;
    }
    
}