LeetCode 312 Burst Balloons (区间dp)

最新推荐文章于 2022-06-05 16:05:32 发布

原创最新推荐文章于 2022-06-05 16:05:32 发布 · 1.3k 阅读

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#leetcode #区间dp

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探讨了如何通过动态规划解决LeetCode上的气球爆破问题，旨在获取最大金币数。文章详细解释了状态转移方程，并提供了一个高效的Java解决方案。

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