實時判斷數據流中的第K大元素

int k = 3;
int[] arr = [4,5,8,2];
KthLargest kthLargest = new KthLargest(3, arr);
kthLargest.add(3);   // returns 4
kthLargest.add(5);   // returns 5
kthLargest.add(10);  // returns 5
kthLargest.add(9);   // returns 8
kthLargest.add(4);   // returns 8

小頂堆實現，保留最大的K個元素。

import java.util.PriorityQueue;

public class KthLargest {

    private PriorityQueue<Integer> queue;

    private int limit;

    public KthLargest(int k, int[] nums) {
        if (k < 0) {
            throw new RuntimeException("k error!");
        }
        limit = k;
        queue = new PriorityQueue<>(k);
        for (int num : nums) {
            add(num);
        }
    }

    public int add(int val) {
        if (queue.size() < limit || queue.peek() == null) {
            queue.add(val);
        } else if (val > queue.peek()) {
            queue.poll();
            queue.add(val);
        }

        return queue.peek();
    }

    /**
     * Your KthLargest object will be instantiated and called as such:
     * KthLargest obj = new KthLargest(k, nums);
     * int param_1 = obj.add(val);
     */
}