[LeetCode]Scramble String

标签：leetcode   java   深搜   备忘录   

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.

深搜+备忘录法

public class Solution {
	Map<String,Boolean> map = new HashMap<>();
    public boolean isScramble(String s1, String s2) {
        if(s1.length()!=s2.length()) return false;
        int len = s1.length();
        if(len == 1) return s1.equals(s2);
        for(int i=1;i<s1.length();i++){
        	if((store(s1.substring(0,i),s2.substring(0,i))&&store(s1.substring(i),s2.substring(i)))||
        		(store(s1.substring(0,i),s2.substring(len-i))&&store(s1.substring(i),s2.substring(0,len-i)))){
        		return true;
        	}
        }
        return false;
    }
    
    private boolean store(String s1, String s2){
    	String str = s1+s2;
    	if(map.containsKey(str)){
    		return map.get(str);
    	}else{
    		boolean foo = isScramble(s1,s2);
    		map.put(str, foo);
    		return foo;
    	}
    }
}

[LeetCode]Scramble String

标签：leetcode   java   深搜   备忘录   

原文地址：http://blog.csdn.net/guorudi/article/details/43195223