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

[java] view
plain copy

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()];    

    }    

}  

原文链接http://blog.csdn.net/crazy__chen/article/details/47377069