leetcode--Scramble String

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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.

public class Solution {
    public boolean isScramble(String s1, String s2) {
       //we have assume that s1 and s2 have the same length and are non-empty
		
		if(s1.equals(s2)) return true;
		
		//check two string contains the same characters
		char[] s1char = s1.toCharArray();
		char[] s2char = s2.toCharArray();
		Arrays.sort(s1char);
		Arrays.sort(s2char);
		for(int i = 0; i < s1.length(); ++i){
			if(s1char[i] != s2char[i])
				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(s1.length() - i)) && isScramble(s1.substring(i), s2.substring(0,s1.length() - i)))
				return true;
		}
		return false;
    }
}

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leetcode--Scramble String

标签：des   blog   java   os   art   for   

原文地址：http://www.cnblogs.com/averillzheng/p/3825778.html