题解 | #最小的K个数#

使用长度为k的优先队列，存储k个最小数字

import java.util.*;


public class Solution {
    /**
     * 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
     *
     *
     * @param input int整型一维数组
     * @param k int整型
     * @return int整型ArrayList
     */
    public ArrayList<Integer> GetLeastNumbers_Solution (int[] input, int k) {
        // write code here
        //robustness


        PriorityQueue<Integer> q =  new PriorityQueue<Integer>((o1,
                o2)->o2.compareTo(o1));

        ArrayList<Integer> list = new ArrayList<Integer>();

        if (k == 0 || input.length == 0)
            return list;

        for (int i = 0; i < k; i++) q.offer(input[i]);

        for (int i = k; i < input.length; i++ ) {
            if (input[i] < q.peek()) {
                q.poll();
                q.offer(input[i]);
            }
        }
        int size = q.size();
        for (int i = 0; i < size; i++)
            list.add(q.poll());

        return list;

    }
}