题目描述
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.
思路：1、对于该问题，首先明白是在判断字符串S2是否为S1的scrambled string，分情况讨论的话，可以刚开始先看是否equals，这样可以判断长度为一的字符串；
2、接下来可以判断字符串的长度是否相同；
3、之后判断是否出现的字符相同，这里新建了一个长度为26的数组，要习惯这种思路，在前面的题中，找出现次数为n的数组中只是出现一次的数字也将32位的值分别存在了长度为32的数组里；
4、对递归的使用，分两种，一种是没有进行左右转换，一种是进行了左右转换；
public class Solution {
public boolean isScramble(String s1, String s2) {
if (s1.equals(s2))
return true;
int[] letters = new int[26];
for (int i = 0; i < s1.length(); i++) {
letters[s1.charAt(i) - 'a']++;
letters[s2.charAt(i) - 'a']--;
}
for (int i = 0; i < 26; i++)
if (letters[i] != 0)
return false;
for (int i = 1; i < s1.length(); i++) {
//当前分割出没有交换
if (isScramble(s1.substring(0, i), s2.substring(0, i))
&& isScramble(s1.substring(i), s2.substring(i))) //因为该函数是返回值为布尔型的，所以可以把其放在条件里面
return true;
//当前分割出交换---左右交换,这两段可以合在一起，但是可读性比较差
if (isScramble(s1.substring(0, i), s2.substring(s2.length() - i))
&& isScramble(s1.substring(i), s2.substring(0, s2.length() - i)))
return true;
}
return false;