LeetCode_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.

　　三维动态规划：  f(i, j, n) = || ((f(i, j, m) && f(i + m, j + m, n - m)) || f(i, j + n - m, m) && f(i + m, j, n - m)) for 1 <= m < n where f(i, j, n) is true iff substring starts at s1[i] and substring starts at s2[j] both with length n are scrambled

class Solution {
public:
    bool isScramble(string s1, string s2) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        int n = s1.size();
        if (s2.size() != n) return false;
        if(s1 == s2 ) return true;

        bool f[n][n][n+1];
        
        for(int i= 0; i< n; i++)
            for(int j = 0; j< n; j++)
            {
                f[i][j][0] = true;//len 0 其实没用，只是为了k 的编程容易而已
                f[i][j][1] = s1[i] == s2[j] ;
            }
            
        for(int len = 2; len <= n; len ++)
            for( int i = 0; i + len -1 < n;i++)
                for(int j = 0; j + len -1 <n;j++)
                {
                    f[i][j][len] = false;
                    for(int k = 1; k< len ; k++){
                        if ( (f[i][j][k] && f[i+k][j+k][len-k]) ||
                                (f[i][j+len-k][k] && f[i+k][j][len-k]) )
                                {
                                     f[i][j][len] = true;
                                     break;
                                }
                    }
                }
                
        return f[0][0][n];
    }
};