LeetCode 312 Burst Balloons (区间dp)

Given n balloons, indexed from 0 to n-1. Each balloon is painted with a number on it represented by arraynums. You are asked to burst all the balloons. If the you burst ballooni you will get nums[left] * nums[i] * nums[right] coins. Hereleft and right are adjacent indices of i. After the burst, theleft and right then becomes adjacent.

Find the maximum coins you can collect by bursting the balloons wisely.

Note:
(1) You may imagine nums[-1] = nums[n] = 1. They are not real therefore you can not burst them.
(2) 0 ≤ n ≤ 500, 0 ≤ nums[i] ≤ 100

Example:

Given [3, 1, 5, 8]

Return 167

    nums = [3,1,5,8] --> [3,5,8] -->   [3,8]   -->  [8]  --> []   coins =  3*1*5      +  3*5*8    +  1*3*8      + 1*8*1   = 167

Credits:
Special thanks to @dietpepsi for adding this problem and creating all test cases.

题目链接：https://leetcode.com/problems/burst-balloons/

题目分析：设dp[i][j]为i到j这段区间所能得到的最大值，状态转移方程为dp[i][j] = max(i < k < j) (dp[i][k] + dp[k][j] + a[i] * a[k] * a[j])
为了方便计算，将数组扩充一头一尾，值均为1，击败了84%

public class Solution {    public int maxCoins(int[] nums) {        int n = nums.length + 2;        int []a = new int[n];        a[0] = 1;        a[n - 1] = 1;        for (int i = 0; i < n - 2; i ++) {            a[i + 1] = nums[i];        }        int [][]dp = new int[n][n];        for (int l = 2; l < n; l ++) {            for (int i = 0; i + l < n; i ++) {                int j = i + l;                for (int k = i + 1; k < j; k ++) {                    dp[i][j] = Math.max(dp[i][j], dp[i][k] + dp[k][j] + a[i] * a[j] * a[k]);                }            }        }        return dp[0][n - 1];    }}