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.
Example 1:
Input: s1 = "great", s2 = "rgeat" Output: true
Example 2:
Input: s1 = "abcde", s2 = "caebd" Output: false题意：给定两个字符串，判断这两个字符串是否互为scramble，scramble的定义是将字符串按照二叉树的形式逐级进行字符拆分，直到不能再拆为止，如果两个字符串互为scramble，那么交换两个叶子节点可以将两个字符串相互转换代码如下：
/**
* @param {string} s1
* @param {string} s2
* @return {boolean}
*/
var isScramble = function(s1, s2) {
if(s1.length!=s2.length) return false;
if(s1==s2) return true;
let str1=s1,str2=s2;
str1=str1.split(‘‘).sort().join(‘‘)
str2=str2.split(‘‘).sort().join(‘‘)
if(str1!=str2) return false;
for(let i=1;i<s1.length;i++){
let s11=s1.substr(0,i);
let s12=s1.substr(i);
let s21=s2.substr(0,i);
let s22=s2.substr(i);
if(isScramble(s11,s21) && isScramble(s12,s22)) return true;
s21=s2.substr(s1.length-i);
s22=s2.substr(0,s1.length-i);
if(isScramble(s11,s21) && isScramble(s12,s22)) return true;
}
return false;
};
原文地址：https://www.cnblogs.com/xingguozhiming/p/10667217.html