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.

思路：

动规。用result[i][j][k]表示s[i..i+k-1]与s[j..j+k-1]是否相等。

代码：

    bool isScramble(string s1, string s2) {
        // IMPORTANT: Please reset any member data you declared, as
        // the same Solution instance will be reused for each test case.
        if(s1.length() != s2.length())
            return false;
        int len = s1.length();
        bool result[len][len][len+1];
        memset(result, false, sizeof(bool)*len*len*(len+1));
        int i,j,k,m;
        for(i = 1; i <= len; i++){
            for(j = 0; j <= len-i; j++){
                for(k = 0; k <= len-1; k++){
                    if(i == 1)
                        result[j][k][i] = (s1[j] == s2[k]);
                    else{
                        for(m = 1 ; m < i; m++){
                            result[j][k][i] = (result[j][k][m] && result[j+m][k+m][i-m])||(result[j][k+i-m][m] && result[j+m][k][i-m]);
                            if(result[j][k][i])
                                break;
                        }                            
                    }
                    cout<<j<<" "<<k<<" "<<i<<": "<<result[j][k][i]<<endl;
                }
            }   
        }
        return result[0][0][len];
    }