2 q294402756 q294402756 于 2015.07.20 23:48 提问

动态规划初始化问题,类似于下面这道题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":
/ \
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".
/ \
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".
/ \
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];


代码中对dp[i][j][0]初始化定义 一般这种动态规划的题目都有初始定义。这种初始定义有什么意义?这道题中的初始化代码有什么意义?一般动态规划的题目怎么确定该如何初始化?求大神来解惑一下,不甚感谢


caozhy   Ds   Rxr 2015.07.21 06:15
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