leetcode:Scramble String问题

Question

题目：
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.

 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];
    }

}

我想问的是 为什么要初始dp[i][j][0]，这样做的意义何在？还有很多动态规划的题目 例如 方格步数 爬楼梯 都要一开始初始化 这样的意义是什么？