本文共 1972 字，大约阅读时间需要 6 分钟。

题目链接：https://leetcode.com/problems/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.

思路：可以用递归来做，也可以用动态规划来做。递归的思路是两个字符串各找一个分割点，例如s1 和 s2使得要么：

1. s1分割点左边的字符串和s2分割点左边的字符串相等或者可以通过交换相等，并且s1右边的子串和s2右子串同样符合此规律

2. s1的左子串和s2的右子串相等或者可以交换使得相等，并且s1的右子串和s2的左子串符合同样的规律

如果不剪枝，会超时，可以通过判断两个字符串的字母是否个数一样来剪枝。

代码如下：

class Solution {public:    bool isScramble(string s1, string s2) {        if(s1 == s2) return true;        if(s1.size() != s2.size() || s1.size() ==1) return false;        int len = s1.size(), num = len;        vector
   
     hash(26, 0);        for(auto val: s1) hash[val-'a']++;        for(auto val: s2) if(hash[val-'a']) hash[val-'a']--, num--;        if(num != 0) return false;        for(int i = 1; i < len; i++)        {            if(isScramble(s1.substr(0, i), s2.substr(0, i)) &&                isScramble(s1.substr(i), s2.substr(i))) return true;            if(isScramble(s1.substr(0, i), s2.substr(len-i)) &&                isScramble(s1.substr(i), s2.substr(0, len-i))) return true;        }        return false;    }};
   

转载地址：http://kslbn.baihongyu.com/