leetcode 703. Kth Largest Element in a Stream（第k大的元素）

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.

Implement KthLargest class:

KthLargest(int k, int[] nums) Initializes the object with the integer k and the stream of integers nums.
int add(int val) Appends the integer val to the stream and returns the element representing the kth largest element in the stream.

Example 1:

Input
[“KthLargest”, “add”, “add”, “add”, “add”, “add”]
[[3, [4, 5, 8, 2]], [3], [5], [10], [9], [4]]
Output
[null, 4, 5, 5, 8, 8]

Explanation
KthLargest kthLargest = new KthLargest(3, [4, 5, 8, 2]);
kthLargest.add(3); // return 4
kthLargest.add(5); // return 5
kthLargest.add(10); // return 5
kthLargest.add(9); // return 8
kthLargest.add(4); // return 8

设计一个类，每次add操作时都能返回第k大的元素。

思路：
如果用最大堆，那么要取k次元素才能取到第k大的，每次都这样取影响效率。
如果用最小堆，最大的元素压在下面，只要保持只有k个元素即可，那每次取的一定是第k大的元素。

class KthLargest {
    PriorityQueue<Integer> heap = new PriorityQueue<Integer>();
    int k = 0;
    
    public KthLargest(int k, int[] nums) {      
        this.k = k;
        for(int num : nums) {
            heap.add(num);
            if(heap.size() > k) heap.poll();
        }
    }
    
    public int add(int val) {
        if(heap.size() == k) {
            if(heap.peek() < val) {
                heap.poll();
                heap.add(val);
            }
        } else {
            heap.add(val);
        }
        return heap.peek();
    }
}