Leetcode: 87. Scramble String

Description

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

• 递归
• 使用一个hash表来判断两个字符串的字符是否相同，若不同，直接返回false
• 由于不知道左右子树从哪里划分，所以循环从1-len来划分左右子树，注意判断两个串的两种情况

代码

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