153.Kth Largest Element in a Stream

题目：

Design a class to find the kth largest element in a stream. Note that it is the kth largest element in the sorted order, not the kth distinct element.

设计一个类来查找流中第k个最大元素。 请注意，它是排序顺序中的第k个最大元素，而不是第k个不同元素。

Your KthLargest class will have a constructor which accepts an integer k and an integer array nums, which contains initial elements from the stream. For each call to the method KthLargest.add, return the element representing the kth largest element in the stream.

你的KthLargest类将有一个构造函数，它接受一个整数k和一个整数数组nums，它包含来自流的初始元素。 对于方法KthLargest.add的每次调用，返回表示流中第k个最大元素的元素。

Example:

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

Note: 
You may assume that nums' length ≥ k-1 and k ≥ 1.

您可以假设nums的长度≥k-1且k≥1。

解答：

方法一：优先级队列（自动排序的特性）

class KthLargest {
    
    final PriorityQueue<Integer> q;
    final int k;

    public KthLargest(int k, int[] nums) {
        this.k=k;
        q=new PriorityQueue<>(k);
        for(int n:nums)
            add(n);
    }
    
    public int add(int val) {
        if(q.size()<k)
            q.offer(val);
        else if(q.peek()<val){
            q.poll();
            q.offer(val);
        }
        return q.peek();
    }
}

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

方法二：快速排序（从大到小）

详解：