LeetCode 703. Kth Largest Element in a Stream（数据流中第K大的元素）

Discuss：www.cnblogs.com/grandyang/p…

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) 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

Constraints:

1 <= k <= 104
0 <= nums.length <= 104
-104 <= nums[i] <= 104
-104 <= val <= 104
At most 104 calls will be made to add.
It is guaranteed that there will be at least k elements in the array when you search for the kth element.

解法一：

使用优先队列，构造一个大小为 k 的小顶堆，堆顶便是我们要的元素。

class KthLargest(k: Int, nums: IntArray) {
    var priorityQueue = PriorityQueue<Int>(k)
    private var k: Int = 1

    init {
        this.k = k
        for (i in nums.indices) {
            if (i < k) {
                priorityQueue.offer(nums[i])
            } else if (priorityQueue.peek() < nums[i]) {
                priorityQueue.poll()
                priorityQueue.offer(nums[i])
            }
        }
    }

    fun add(value: Int): Int {
        if (priorityQueue.size < k) {
            priorityQueue.offer(value)
        } else if (priorityQueue.peek() < value) {
            priorityQueue.poll()
            priorityQueue.offer(value)
        }
        return priorityQueue.peek()
    }
}