87. Scramble String（js）

最新推荐文章于 2025-08-13 13:14:14 发布

转载最新推荐文章于 2025-08-13 13:14:14 发布 · 84 阅读

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原文链接：http://www.cnblogs.com/xingguozhiming/p/10667217.html

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#数据结构与算法

本文介绍了一种判断两个字符串是否互为Scramble的方法。Scramble定义为将字符串按二叉树形式逐级拆分，通过交换叶子节点实现字符串转换。文章提供了具体的算法实现，并举例说明。

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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;
};