leetcode--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.

题意：如图，判断两个字符串是否是scrambled关系。

分类：动态规划，字符串

解法1：三维动态规划。简单的说，就是s1和s2是scramble的话，那么必然存在一个在s1上的长度l1，将s1分成s11和s12两段，同样有s21和s22。
那么要么s11和s21是scramble的并且s12和s22是scramble的(条件1)；
要么s11和s22是scramble的并且s12和s21是scramble的(条件2)。

思路是建立一个boolean flag[i][j][len]，表明s1从i开始，s2从j开始，len长度的字符串，两者是不是scrambled关系。所以我们求的是flag[0][0][len]

根据上面的说法，要判断flag[i][j][len]是否为true，我们有两个判断依据

首先，我们尝试将s1分成两部分，因为起始序号是i,结束序号是i+len-1

假设我们从i开始，k长度把s1分成s11,s12，那么0<=k<=len

所以我们要判断res[i][j][k]&&res[i+k][j+k][len-k]是否为true,如果是，则满足条件1，

另外，我们还可以根据条件2

判断res[i][j+len-k][k]&&res[i+k][j][len-k]是否为true

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<s1.length();i++){  
            for(int j=0;j<s2.length();j++){  
                res[i][j][1] = s1.charAt(i)==s2.charAt(j);  
            }  
        }  
        for(int len=2;len<=s1.length();len++){  
            for(int i=0;i<s1.length()-len+1;i++){  
                for(int j=0;j<s2.length()-len+1;j++)  {  
                    for(int k=1;k<len;k++){  
                        res[i][j][len] |= res[i][j][k]&&res[i+k][j+k][len-k] || res[i][j+len-k][k]&&res[i+k][j][len-k];  
                    }  
                }  
            }  
        }  
        return res[0][0][s1.length()];  
    }  
}