【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+)));
         //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-;i>=;i--)
         {
             for(int j=n-;j>=;j--)
             {
                 for(int k=;k<=n-max(i,j);k++)
                 {
                     if(s1.substr(i,k)==s2.substr(j,k))
                     {
                         dp[i][j][k]=true;
                     }
                     else
                     {
                         for(int l=;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[][][n];
     }
 };