Scrambling 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.

 public boolean isScramble(String s1, String s2) {   
   int len = s1.length();   
   if (len != s2.length()) return false;   
   if (s1.equals(s2)) return true;   
   // a table of matches   
   // T[i][j][k] = true iff s2.substring(j,j+k+1) is a scambled string of s1.substring(i,i+k+1)   
   boolean[][][] scrambled = new boolean[len][len][len];   
   for (int i=0; i < len; ++i) {   
    for (int j=0; j < len; ++j) {   
     scrambled[i][j][0] = (s1.charAt(i) == s2.charAt(j));   
    }   
   }   
   // dynamically fill up the table   
   for (int k=1; k < len; ++k) { // k: length   
    for (int i=0; i < len - k; ++i) { // i: index in s1   
     for (int j=0; j < len - k; ++j) { // j: index in s2   
      scrambled[i][j][k] = false;   
      for (int p=0; p < k; ++p) { // p: split into [0..p] and [p+1..k]   
       if ((scrambled[i][j][p] && scrambled[i+p+1][j+p+1][k-p-1])   
         || (scrambled[i][j+k-p][p] && scrambled[i+p+1][j][k-p-1])) {   
        scrambled[i][j][k] = true;   
        break;   
       }   
      }   
     }   
    }   
   }   
   return scrambled[0][0][len-1];   
  }  

这个问题的解释详见:http://n00tc0d3r.blogspot.com/2013/05/scramble-string.html