leetcode.87. Scramble String

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本文介绍了一种判断两个字符串是否可以通过子串交换操作相互转换的算法。通过递归分割字符串成子串，并检查子串是否能通过特定方式匹配来实现。文章详细解释了算法的工作原理及其实现细节。

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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.

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

    }
};