leetCode 87. 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.

大意是字符串局部镜像转换的判断

规律在于只要在s1的任意位置一分为二镜像交换位置能够变成s2，则返回true，否则返回false。当然s1一分为二后还要继续分，这时候用递归就好了。

为了加快速度，还需要剪枝，这里是把字符串转换为字符数组然后排序实现，用map计数也可以

public static boolean isScramble(String s1, String s2) {
if(s1.equals(s2))return true;
if(s1.length()!=s2.length())return false;

char[]A = s1.toCharArray();
Arrays.sort(A);
String temp1 = String.valueOf(A);

A = s2.toCharArray();
Arrays.sort(A);
String temp2 = String.valueOf(A);
if(!temp1.equals(temp2)){
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))
||
isScramble(s1.substring(0,i),s2.substring(s2.length()-i))&&
isScramble(s1.substring(i),s2.substring(0,s2.length()-i))){
return true;
}
}

return false;
}