No312. Burst Balloons

最新推荐文章于 2022-02-16 22:42:53 发布

森宝宝

最新推荐文章于 2022-02-16 22:42:53 发布

阅读量252

点赞数

CC 4.0 BY-SA版权

文章标签： LeetCode dp

本文链接：https://blog.youkuaiyun.com/qiaosen20/article/details/73498445

通过动态规划解决爆破气球问题，目标是在遵循特定规则的情况下获得最大金币数。给定数组表示每个气球上的数值，爆破时根据相邻气球数值计算金币数。

一、题目描述

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

二、主要思路

动态规划解题，首先在vector的首尾各插入一个1，二维数组dp用于记录从下标为i到j的最大值，对应的状态转移方程为
dp[l][r]=max(dp[l][r],(dp[l][i]+dp[i][r]+nums[l]*nums[i]*nums[r]))，即在l、r和len取不同值的时候，dp[l][r]取
dp[l][i]+dp[i][r]+nums[l]*nums[i]*nums[r]的最大值，表示nums[l]、nums[i]和nums[r]是最后相乘的三个数，他们彼此之间数取最大值，即dp[l][i]和dp[i][r]。

最后返回从坐标0到n-1的dp值。

三、代码实现

class Solution {
public:
    int maxCoins(vector<int>& nums) {
        nums.insert(nums.begin(),1);
        nums.push_back(1);
        int n=nums.size();
        vector<vector<int>>dp(n,vector<int>(n));
        for(int len=3;len<=n;len++){
            for(int l=0;l<=n-len;l++){
                int r=l+len-1;
                for(int i=l+1;i<r;i++){
                    dp[l][r]=max(dp[l][r],(dp[l][i]+dp[i][r]+nums[l]*nums[i]*nums[r]));
                }
            }
        }
        return dp[0][n-1];
    }
};