https://leetcode.com/problems/total-cost-to-hire-k-workers/description/
2462. Total Cost to Hire K Workers
給你一個陣列costs，costs[i] 表示第 i 個應徵者的期望薪資，我們要開一個會議
每次從最左邊和最右的應徵者開始挑選 candidates 個人，並從中雇用最便宜的應徵
者（成本相同的話靠左的人優先），剩下的人進入下一輪會議，公司需要挑選出 k 個人，求出雇用的最小成本。
Example 1:
Input: costs = [17,12,10,2,7,2,11,20,8], k = 3, candidates = 4
Output: 11
Explanation: We hire 3 workers in total. The total cost is initially 0.
- In the first hiring round we choose the worker from
[17,12,10,2,7,2,11,20,8]. The lowest cost is 2, and we break the tie by the
smallest index, which is 3. The total cost = 0 + 2 = 2.
- In the second hiring round we choose the worker from
[17,12,10,7,2,11,20,8]. The lowest cost is 2 (index 4). The total cost = 2 +
2 = 4.
- In the third hiring round we choose the worker from [17,12,10,7,11,20,8].
The lowest cost is 7 (index 3). The total cost = 4 + 7 = 11. Notice that the
worker with index 3 was common in the first and last four workers.
The total hiring cost is 11.
思路：
1.照題目需求，每次都從最左邊和最右邊挑選 candidates 個人並放進 Min Heap 裡
按 costs 排序，相同就比較索引，這樣每次拿出來的就會是最優解，如果取了左邊的
人那下次就要先從左邊補人，右邊反之。
（原始的解法太醜了我就不貼了）
2.上面是比較直觀的解，還可以做一些優化，開頭可以直接判斷 candidates 是否可以
直接包含全部 costs，這樣只要直接排序後取 k 個就是最佳解，不必一直維護heap。
3.可以用兩個 Heap 來保存左半邊和右半邊，左半邊的 Heap 優先取出，可以免除一些
繁瑣的判斷。
Java Code：