Scramble String

Problem:

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.

参考了CSDN博客 http://blog.csdn.net/pickless/article/details/11501443。

在此对作者表示感谢。

Solution1：枚举分割点i，分别判断isScramble(s1左，s2左)&&isScramble(s1右，s2右)和isScramble(s1左，s2右)&&isScramble(s1右，s2左)是否成立。

Solution2：动态规划。
定义F[i][j][k]，
当F[i][j][k]=true时，表示S1[i..i+k-1] == S2[j..j+k-1]。
用白话表达：F[i][j][k]记录的是S1从i开始k个字符与S2从j开始k个字符是否为Scramble String。

最简单的情况为k=1时，即S1[i]与S2[j]是否为Scramble String。
因此F[i][j][1] = S1[i] == S2[j]。

当K=2时，
F[i][j][2] = (F[i][j][1] && F[i+1][j+1][1]) || (F[i][j+1][1] && F[i+1][j][1])。
F[i][j][1] && F[i+1][j+1][1]表达的是 S1[i] == S2[j] && S1[i+1]==S2[j+1]（例如：“AC”和“AC”），如果这个条件满足，那么S1[i..i+1]与S2[j..j+1]自然为Scramble String，即F[i][j][2] = true。
F[i][j+1][1] && F[i+1][j][1]表达的是S1[i+1] == S2[j] && S1[i] == S2[j + 1] （例如： “AB”和“BA”），同样如果该条件满足，F[i][j][2] = true。


当K为更大的数时，同递归算法一样，我们需要枚举分割点，假设左边长度为l，即S[i..i+l-1]，右边长度为k-l，即S[i+l..i+k-1]。
同样存在两种情况，S1左 = S2左 && S1右 = S2右 或者 S1左 = S2右 && S1右 = S2左。

Solution1:

public class Solution {
    public boolean isScramble(String s1, String s2) {
        if(s1.equals(s2))
         return true;
        char[] ch1 = s1.toCharArray();
        char[] ch2 = s2.toCharArray();
        Arrays.sort(ch1);
        Arrays.sort(ch2);
        if(!new String(ch1).equals(new String(ch2)))
         return false;
        for(int i=1;i<s1.length();i++)
        {
         if(isScramble(s1.substring(0,i), s2.substring(0,i))&&isScramble(s1.substring(i,s1.length()), s2.substring(i,s2.length())))
         return true;
         if(isScramble(s1.substring(0,i),s2.substring(s2.length()-i,s2.length()))&&isScramble(s1.substring(i,s1.length()), s2.substring(0,s2.length()-i)))
         return true;
        }
        return false;
    }

}

Solution2：

public class Solution {
public boolean isScramble(String s1, String s2) {
    if(s1.length()!=s2.length())
        return false;
    int len = s1.length();
     boolean[][][] scr = new boolean[len][len][len+1];

    for(int k=1;k<=len;k++)
    {
        for(int i=0;i<=len-k;i++)
        {
             for(int j=0;j<=len-k;j++)
            {
                if(k==1)
                {
                    scr[i][j][k] = s1.charAt(i)==s2.charAt(j);
                }
                else 
                {
                   for(int l=1;l<k;l++)
                   {
                       if((scr[i][j][l]&&scr[i+l][j+l][k-l])||(scr[i][j+l][k-l]&&scr[i+k-l][j][l]))
                       {
                           scr[i][j][k] = true;
                           break;
                       }
                   }
                 }
            }
        }
     }
     return scr[0][0][len];
}
}