lintcode: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 ifs2 is a scrambled string ofs1.

public class Solution {    /**     * @param s1 A string     * @param s2 Another string     * @return whether s2 is a scrambled string of s1     */    public boolean isScramble(String s1, String s2) {        // Write your code here        int n = s1.length();        if (s2.length() != n)             return false;                    boolean dp[][][] = new boolean[n][n][n];                // case: length is 1        for (int i=0; i<n; i++)            for (int j=0; j<n; j++)                dp[i][j][0] = s1.charAt(i) == s2.charAt(j);                                // case: length is 2....n        for (int l=1; l<n; l++)        {            for (int i=0; i+l<n; i++)            {                for (int j=0; j+l<n; j++)                {                    for (int k=0; k<l; k++)                    {                        if ((dp[i][j][k] && dp[i+k+1][j+k+1][l-1-k])                            || (dp[i][j+l-k][k] && dp[i+k+1][j][l-1-k]))                            dp[i][j][l] = true;                    }                }            }        }                return dp[0][0][n-1];    }        }