【Leetcode】Scramble String - 最难动态规划 3维

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.

Java：

http://blog.csdn.net/linhuanmars/article/details/24506703

s1 s2长度一样 才能比 那其实都用一个的长度当length是不是也行？

public class Solution {
    public boolean isScramble(String s1, String s2) {
        if(s1==null||s2==null||s1.length()!=s2.length())return false;
        if(s1.length()==0) return true;
        boolean[][][] res= new boolean[s1.length()][s2.length()][s1.length()+1];
        for(int i=0;i