87. Scramble String

Given a strings1, we may represent it as a binary tree by partitioning it to two non-empty substrings recursively.

Below is one possible representation ofs1="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 s1and s2 of the same length, determine if s2 is a scrambled string of s1.

S: 记忆化搜索 O(n^4)?

记忆化搜索主要目的是存储搜索中间结果避免重复计算，这里中间结果是两个string比较的结果，用hashmap存储

class Solution {
public:
    bool isScramble(string s1, string s2) {
        if (s1.size() != s2.size()) {
            return false;
        }
        if (s1 == s2) {
            return true;
        }
        string key = s1 + s2;
        int n = s1.size();
        if (dp.find(key) != dp.end()) {
            return dp[key];
        }
        for (int i = 1; i < n; ++i) {
            if ((isScramble(s1.substr(0, i), s2.substr(0, i)) && 
                isScramble(s1.substr(i, n - i), s2.substr(i, n - i))) ||
                (isScramble(s1.substr(0, i), s2.substr(n - i, i)) &&
                isScramble(s1.substr(i, n - i), s2.substr(0, n - i)))) {
                dp[key] = true;
                return dp[key];
            }
        }
        dp[key] = false;
        return dp[key];
    }

private:
    unordered_map<string, bool> dp;        
};