题目

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":

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    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.

方法

该题目用到了三维动态规划。到眼下为止没有理解深入。 參考了：http://blog.csdn.net/jiyanfeng1/article/details/8620224              http://www.cnblogs.com/remlostime/archive/2012/11/19/2778108.html 
    public boolean isScramble(String s1, String s2) {
    	if (s1.equals(s2)) {
    		return true;
    	}
    	int len1 = s1.length();
    	int len2 = s2.length();
    	boolean[][][] scrambled = new boolean[len1][len2][len1 + 1];
    	for (int i = 0; i < len1; i++) {
    		for (int j = 0; j < len2; j++) {
    			scrambled[i][j][0] = true;
    			scrambled[i][j][1] = (s1.charAt(i) == s2.charAt(j));
    		}
    	}
    	
    	for (int i = len1 - 1; i >= 0; i--) {
    		for (int j = len2 - 1; j >= 0; j--) {
    			for (int n = 2; n <= Math.min(len1 - i, len2 - j ); n++) {
    				for (int m = 1; m < n; m++) {
                        scrambled[i][j][n] |= scrambled[i][j][m] && scrambled[i + m][j + m][n - m] ||  
                                scrambled[i][j + n - m][m] && scrambled[i + m][j][n - m];  
                        if(scrambled[i][j][n])  break;  
    				}
    			}
    		}
    	}
    	return scrambled[0][0][len1];
    }




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