[hard]312. Burst Balloons

最新推荐文章于 2025-12-02 18:04:47 发布

原创最新推荐文章于 2025-12-02 18:04:47 发布 · 226 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#leetcode #DP

本篇介绍了一个算法问题——通过智慧地爆破一系列带有数值的气球来获得最大金币数。提供了详细的解析过程及示例，利用动态规划求解最优解。

312. Burst Balloons

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.

Solution:

F[i,j] ------------------------------------- the max score you got when you had bursted from i + 1 to j - 1

So, We have

f[i][k] = max(f[i][k], f[i][j] + f[j][k] + nums[i] * nums[j] * nums[k]); where j is i + 1 to k - 1

class Solution {
public:
    int maxCoins(vector<int>& nums) {
        if (nums.size() == 0)
            return 0;
        if (nums.size() == 1)
            return nums[0];

        nums.insert(nums.begin(), 1);
        nums.push_back(1);
        
        int f[nums.size()][nums.size()]; memset(f, 0, sizeof(f));

        for (int len = 2; len < nums.size(); ++ len)
            for (int i = 0; i < nums.size() - len; ++ i) {
                int k = i + len;
                for (int j = i + 1; j < k; ++ j)
                    f[i][k] = max(f[i][k], f[i][j] + f[j][k] + nums[i] * nums[j] * nums[k]);
            }

        return f[0][nums.size() - 1];
    }
};