703. Kth Largest Element in a Stream - Easy

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.

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.

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.

 

建立一个min heap，只包含k个最大的元素。

constructor把nums各项放入min heap，只保留k个最大数。add函数先把当前val放入min heap，如果heap的size > k就移出堆顶最小元素。最后返回min heap的堆顶元素

*注意记得k的处理

时间O(NlogK)，空间O(K)

class KthLargest {
    PriorityQueue<Integer> minHeap;
    int k;

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