LeetCode 0312 - Burst Balloons

Reck Zhang

发布于 2021-08-11 11:14:32

24100

代码可运行

文章被收录于专栏：Reck ZhangReck Zhang

运行总次数：0

代码可运行

Burst Balloons

Desicription

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:

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

Example:

Input: [3,1,5,8]
Output: 167 
Explanation: 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

Solution

class Solution {
public:
    int maxCoins(std::vector<int>& nums) {
        auto balloons = std::vector<int>(nums.size() + 2);
        balloons[0] = balloons[balloons.size() - 1] = 1;
        for(int i = 1; i < balloons.size() - 1; i++) {
            balloons[i] = nums[i - 1];
        }
        
        auto dp = std::vector<std::vector<int>>(balloons.size(), std::vector<int>(balloons.size(), 0));
        for(int length = 2; length < balloons.size(); length++) {
            for(int left = 0; left + length < balloons.size(); left++) {
                int right = left + length;
                for(int i = left + 1; i < right; i++) {
                    dp[left][right] = std::max(dp[left][right], balloons[left] * balloons[i] * balloons[right] + dp[left][i] + dp[i][right]);
                }
            }
        }
        
        return dp[0][balloons.size() -1];
    }
};

本文参与腾讯云自媒体同步曝光计划，分享自作者个人站点/博客。

原始发表：2019-05-13，如有侵权请联系 cloudcommunity@tencent.com 删除

each