LeetCode #312 - Burst Balloons

最新推荐文章于 2022-03-16 07:17:43 发布

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最新推荐文章于 2022-03-16 07:17:43 发布

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解决一个经典的动态规划问题——气球爆破问题，通过合理的策略爆破气球以获得最大硬币数。介绍如何利用动态规划算法求解，并提供详细的代码实现。

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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 rightare 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

给定n个气球，每个气球代表一个数值，当戳破气球i就会获得nums[left]*nums[i]*nums[right]个硬币，求戳破所有的气球可能获得的最大的硬币数。自然是利用动态规划，但是递推关系并不好找，dp[i][j]表示戳破从气球i到气球j的所有气球得到的最大硬币数，那么对于i<=k<=j，有递推公式dp[i][j]=max(dp[i][j],nums[k]*nums[i-1]*nums[j+1]+dp[i][k-1]+dp[k+1][j]),由于是按照对角线的顺序遍历动态规划的矩阵，所以递推的循环写法稍有改变。

class Solution {
public:
    int maxCoins(vector<int>& nums) {
        int n=nums.size();
        nums.insert(nums.begin(),1);
        nums.insert(nums.end(),1);
        vector<vector<int>> dp(n+2,vector<int>(n+2,0));
        for(int count=1;count<=n;count++)
        {
            int i=1;
            for(int j=count;j<=n;j++)
            {
                for(int k=i;k<=j;k++)
                    dp[i][j]=max(dp[i][j],nums[k]*nums[i-1]*nums[j+1]+dp[i][k-1]+dp[k+1][j]);
                i++;
            }
        }
        return dp[1][n];
    }
};