LeetCode 502. IPO (Java版; Hard)-蒲公英云

LeetCode 502. IPO (Java版; Hard)

小灰灰 2023-07-18 06:24 89阅读 0赞

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LeetCode 502. IPO (Java版; Hard)

题目描述

Suppose LeetCode will start its IPO soon. In order to sell a good price of its shares to Venture Capital,
LeetCode would like to work on some projects to increase its capital before the IPO. Since it has limited
resources, it can only finish at most k distinct projects before the IPO. Help LeetCode design the best 
way to maximize its total capital after finishing at most k distinct projects.
You are given several projects. For each project i, it has a pure profit Pi and a minimum capital of Ci is
needed to start the corresponding project. Initially, you have W capital. When you finish a project, you 
will obtain its pure profit and the profit will be added to your total capital.
To sum up, pick a list of at most k distinct projects from given projects to maximize your final capital, 
and output your final maximized capital.
Example 1:
Input: k=2, W=0, Profits=[1,2,3], Capital=[0,1,1].
Output: 4
Explanation: Since your initial capital is 0, you can only start the project indexed 0.
             After finishing it you will obtain profit 1 and your capital becomes 1.
             With capital 1, you can either start the project indexed 1 or the project indexed 2.
             Since you can choose at most 2 projects, you need to finish the project indexed 2 to get the maximum capital.
             Therefore, output the final maximized capital, which is 0 + 1 + 3 = 4.
Note:
You may assume all numbers in the input are non-negative integers.
The length of Profits array and Capital array will not exceed 50,000.
The answer is guaranteed to fit in a 32-bit signed integer.

第一次做; 核心: 1) 成本的小根堆, 利润的大根堆; 将资金可承受的项目放入按照利润排行的大根堆中 2) 纯利润是指: 去掉成本后能挣的钱 3) 使用lambda表达式写比较器很方便

/* 以成本为参考, 将项目放入小根堆 以利润为参考, 将项目放入大根堆 这样分别考虑的缺点是, 大根堆的堆顶元素成本可能大于W.... 先选出W能承受的项目, 再将这些项目加入大根堆, 这样大根堆中的项目就是W能支付的了, 并且堆顶的利润最高 */
class Solution { 
    class Project{ 
        int profit;
        int capital;
        Project(int profit, int capital){ 
            this.profit = profit;
            this.capital = capital;
        }
    }
    public int findMaximizedCapital(int k, int W, int[] Profits, int[] Capital) { 
        int n = Profits.length;
        PriorityQueue<Project> minHeap = new PriorityQueue<>((a,b)->a.capital-b.capital);//成本的小根堆
        PriorityQueue<Project> maxHeap = new PriorityQueue<>((a,b)->b.profit-a.profit);//利润的大根堆
        for(int i=0; i<n; i++){ 
            minHeap.add(new Project(Profits[i], Capital[i]));
        }
        for(int i=0; i<k; i++){ 
            while(!minHeap.isEmpty() && W >= minHeap.peek().capital){ 
                maxHeap.add(minHeap.poll());
            }
            if(maxHeap.isEmpty()){ 
                break;
            }
            W = W + maxHeap.poll().profit;
        }
        return W;
    }
}