【ATT】Scramble String

Q:

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.

A: 三维动态规划。

最初的想法是用递归去做，发现有很多重叠的子问题。因此改用动态规划去存储中间状态。

dp[i][j][k]：初始位置为i，长度为k的s1子串和初始位置为j，长度为k的s2子串是否scramble的。

边界条件是dp[i][j][1].

k>=2时，

计算dp[i][j][k]时，对于任意的l<k,>0,满足子串s1[i~,i+l-1]和s2[j,j+l-1]scramble，同时剩余的两部分也scramble；或者s1[i+k-l,i+k-1]和s2[j,j+l-1]scramble,同时剩余的两部分也scramble。！！注意，不要只考虑一种情况。

复杂度O(n^4).

    bool isScramble(string s1, string s2) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        if(s1.empty()&&s2.empty())
            return true;
            
        int n = s1.size();
        
        vector<vector<vector<bool>>> dp(n,vector<vector<bool>> (n,vector<bool>(n+1,false)) );
        
        int i,j,k,l;
        for(i=0;i<n;i++)
            for(j=0;j<n;j++)
                dp[i][j][1] = s1[i]==s2[j];
                
        for(k=2;k<=n;k++)
        {
            for(i=0;i<=n-k;i++)
            {
                for(j=0;j<=n-k;j++)
                {
                    for(l=1;l<k;l++)
                    {
                        dp[i][j][k] = (dp[i][j][l]&&dp[i+l][j+l][k-l])||(dp[i][j+k-l][l]&&dp[i+l][j][k-l]);
                        if(dp[i][j][k])
                            break;
                    }
                }
            }
        }
        
        return dp[0][0][n];
        
    }