[leetcode]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.

 分治法

这道题的难度不大，但是题意理解起来比较难，最近几天木有写代码的状态 ，不多说了，明天把最后一道题解决掉，然后开始第二遍好好整理一下

代码看起来没啥难度：

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;
	}