一、题目描述

There are n engineers numbered from 1 to n and two arrays: speed and efficiency, where speed[i] and efficiency[i] represent the speed and efficiency for the i-th engineer respectively. Return the maximum performance of a team composed of at most k engineers, since the answer can be a huge number, return this modulo 10^9 + 7.

The performance of a team is the sum of their engineers’ speeds multiplied by the minimum efficiency among their engineers.

输入：n = 6, speed = [2,10,3,1,5,8], efficiency = [5,4,3,9,7,2], k = 2
输出：60
解释：
我们选择工程师 2（speed=10 且 efficiency=4）和工程师 5（speed=5 且 efficiency=7）。
他们的团队表现值为 performance = (10 + 5) * min(4, 7) = 60 。

题意：当两种方案的 efficiency 相等时，speed 之和更大的方案显然更优。问题转化成了如何快速求出每种 efficiency 的最优方案。

算法

• 每个人按照效率 e 进行降序排列。
• 在数组种选取 k 个最大的 speed。

public int maxPerformance(int n, int[] speed, int[] efficiency, int k) {
  int[][] pair = new int[n][2];
  for (int i = 0; i < n; i++) {
      pair[i][0] = speed[i];
      pair[i][1] = efficiency[i];
  }
  Arrays.sort(pair, (e1, e2) -> e2[1] - e1[1]);
  
  PriorityQueue<Integer> pQ = new PriorityQueue<>();
  long max = 0, sd = 0;
  for (int i = 0; i < n; i++) {
      pQ.add(pair[i][0]);
      sd += pair[i][0];
      if (pQ.size() > k) {
          sd -= pQ.poll();
      }
      max = Math.max(max, sd * pair[i][1]);
  }
  return (int)(max % ((int)1e9 + 7));
}

复杂度分析

• 时间复杂度： $O(n log n)$，
• 空间复杂度： $O(k)$，