【leetcode】Scramble String

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.

 

 

1. 如果两个substring相等的话，则为true
2. 如果两个substring中间某一个点，左边的substrings为scramble string， 同时右边的substrings也为scramble string，则为true
3. 如果两个substring中间某一个点，s1左边的substring和s2右边的substring为scramble string, 同时s1右边substring和s2左边的substring也为scramble string，则为true

 

 

class Solution {
public:
    bool isScramble(string s1, string s2) {
       
        int n=s1.length();
        vector<vector<vector<bool>>> dp(n,vector<vector<bool>>(n,vector<bool>(n+1)));
        //dp[i][j][k] represent whether s1[i,i+1,...,i+k-1] and  s2[j,j+1,...,j+k-1] is scramble
       
       
        for(int i=n-1;i>=0;i--)
        {
            for(int j=n-1;j>=0;j--)
            {
                for(int k=1;k<=n-max(i,j);k++)
                {
                    if(s1.substr(i,k)==s2.substr(j,k))
                    {
                        dp[i][j][k]=true;
                    }
                    else
                    {
                        for(int l=1;l<k;l++)
                        {
                            if(dp[i][j][l]&&dp[i+l][j+l][k-l]||dp[i][j+k-l][l]&&dp[i+l][j][k-l])
                            {
                                dp[i][j][k]=true;
                                break;
                            }
                        }
               
                    }
                }
            }
 
        }
        return dp[0][0][n];
    }
};