[LeetCode] 1962. Remove Stones to Minimize the Total

You are given a 0-indexed integer array piles, where piles[i] represents the number of stones in the ith pile, and an integer k. You should apply the following operation exactly k times:
Choose any piles[i] and remove floor(piles[i] / 2) stones from it.
Notice that you can apply the operation on the same pile more than once.

Return the minimum possible total number of stones remaining after applying the k operations.

floor(x) is the greatest integer that is smaller than or equal to x (i.e., rounds x down).

Example 1:
Input: piles = [5,4,9], k = 2
Output: 12
Explanation: Steps of a possible scenario are:

• Apply the operation on pile 2. The resulting piles are [5,4,5].
• Apply the operation on pile 0. The resulting piles are [3,4,5].
  The total number of stones in [3,4,5] is 12.

Example 2:
Input: piles = [4,3,6,7], k = 3
Output: 12
Explanation: Steps of a possible scenario are:

• Apply the operation on pile 2. The resulting piles are [4,3,3,7].
• Apply the operation on pile 3. The resulting piles are [4,3,3,4].
• Apply the operation on pile 0. The resulting piles are [2,3,3,4].
  The total number of stones in [2,3,3,4] is 12.

Constraints:
1 <= piles.length <= 105
1 <= piles[i] <= 104
1 <= k <= 105

移除石子使总数最小。

给你一个整数数组 piles ，数组 下标从 0 开始 ，其中 piles[i] 表示第 i 堆石子中的石子数量。另给你一个整数 k ，请你执行下述操作 恰好 k 次：
选出任一石子堆 piles[i] ，并从中 移除 floor(piles[i] / 2) 颗石子。
注意：你可以对 同一堆 石子多次执行此操作。
返回执行 k 次操作后，剩下石子的 最小 总数。
floor(x) 为 小于 或 等于 x 的 最大 整数。（即，对 x 向下取整）。
来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/remove-stones-to-minimize-the-total
著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。

思路

思路是贪心。这里我们需要借助一个优先队列构建的最大堆，将每一堆的石子数加入最大堆，并同时记录石子的总数，记为 sum。每次弹出堆顶的元素并除以 2，同时记得把减去的部分从 sum 中减去。这样最后 sum 就是全局剩下的石子总数。

复杂度

时间O(nlogn)
空间O(n)

代码

Java实现

class Solution {
    public int minStoneSum(int[] piles, int k) {
        PriorityQueue<Integer> queue = new PriorityQueue<>((a, b) -> b - a);
        int sum = 0;
        for (int p : piles) {
            sum += p;
            queue.offer(p);
        }

        int res = 0;
        while (!queue.isEmpty() && k != 0) {
            int top = queue.poll();
            int remove = top / 2;
            res += remove;
            queue.offer(top - remove);
            k--;
        }
        return sum - res;
    }
}