scramble-string Java code

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.

import java.util.*;
public class Solution {
public boolean isScramble(String s1, String s2) {
int len1 = s1.length();
int len2 = s2.length();
if (len1 != len2)
return false;
if (len1 == 1)
return s1.equals(s2);
//剪枝:若排序后不不相等则必定不满足条件
char[] chars1 = new char[len1];
s1.getChars(0, len1, chars1, 0);
Arrays.sort(chars1);
char[] chars2 = new char[len1];
s2.getChars(0, len2, chars2, 0);
Arrays.sort(chars2);
if (!(new String(chars1).equals(new String(chars2))))
return false;

for (int i = 1; i < len1; i++) {
String s1left = s1.substring(0, i);
String s1right = s1.substring(i, len1);
String s2left = s2.substring(0, i);
String s2right = s2.substring(i, len1);

//在当前分割处没有交换
if (isScramble(s1left, s2left) && isScramble(s1right, s2right))
return true;
//当前分割处左右交换
s2right = s2.substring(len1 - i, len1);
s2left = s2.substring(0, len1 - i);

if (isScramble(s1left, s2right) && isScramble(s1right, s2left))
return true;
}
return false;
}
}