[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：

 1 public class Solution {
 2     public boolean isScramble(String s1,String s2){
 3         if(s1.length() != s2.length()) return false;
 4         if(s1.length() <= 1 && s1.equals(s2)) return true;
 5         if(s1.length() == 2){
 6             if(s1.equals(s2) ) return true;
 7             if(s1.charAt(0) == s2.charAt(1) && s1.charAt(1) == s2.charAt(0)) return true;
 8             return false;
 9         }
10         char[] s1Array = s1.toCharArray();
11         char[] s2Array = s2.toCharArray();
12         Arrays.sort(s2Array);
13         Arrays.sort(s1Array);
14         for(int i = 0; i < s1.length(); i++){
15             if(s1Array[i] != s2Array[i])
16                 return false;
17         }
18         boolean result = false;
19         for(int i = 1; i < s1.length();i++){
20             String s1pre = s1.substring(0,i);
21             String s1suf = s1.substring(i);
22             String s2pre = s2.substring(0, i);
23             String s2suf = s2.substring(i);
24             result = isScramble(s1pre, s2pre) && isScramble(s1suf, s2suf);
25             if(!result){
26                 String s3pre = s2.substring(0, s1.length() - i);
27                 String s3suf = s2.substring(s1.length() - i);
28                 result = isScramble(s1pre, s3suf) && isScramble(s1suf, s3pre);
29             }
30             if(result) return true;
31         }
32         return result;
33     }
34     
35 }

 

posted on 2014-09-01 23:38  喵星人与汪星人  阅读(146)  评论(0)  编辑  收藏  举报