703 Kth Largest Element in a Stream

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

Your KthLargest class will have a constructor which accepts an integerkand an integer arraynums, which contains initial elements from the stream. For each call to the methodKthLargest.add, return the element representing the kth largest element in the stream.

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 andk≥ 1.

这题好像是真题，在面FB时就被当作follow up，估计是我脑残一上来就给quick select。嘛，这题其实比较简单。用的是heap，因为我们求largest，所以用min heap。

class KthLargest {
    Queue<Integer> pq;
    int k;
    public KthLargest(int k, int[] nums) { // Constructor
        pq = new PriorityQueue<>();
        this.k = k;
        if (nums == null || nums.length == 0) {
            return;
        }

        for (int i = 0; i < nums.length; i++) { // 其实这里可以重复利用add的代码
            pq.offer(nums[i]);
            if (pq.size() > k) {
                pq.poll();
            }
        }        
    }

    public int add(int val) {
        pq.offer(val);
        if (pq.size() > k) {            
            pq.poll();
        }
        return pq.peek();
    }
}

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