[Leetcode]703. Kth Largest Element in a Stream 数据流中的第 K 大元素

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) 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.

题意

设计一个找到数据流中第 k 大元素的类（class）。注意是排序后的第 k 大元素，不是第 k 个不同的元素。

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思路

1. 先试着对给定的an integer k and an integer array nums进行一下构造函数constructoer的设计使得其包含数据流最初这些给定参数
2. 每次从数据流新加入的data ，就跟堆顶元素比较一下：比堆顶元素小，直接忽略；比堆顶元素大，就堆顶元素让位。而该堆mantain了k size个最大元素，所以每次返回的堆顶元素，就是第K 大元素

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代码

/*
Time：O(n * logk) 假设数据个数为n对每个数据，最坏情况下，每次都打掉堆顶元素，priorityqueue内部又得重新排序，ksize的priorityqueue的排序时间为logk, 所以总时间是 n 乘以 logk
Space: O(k) 额外开了 k size的priorityqueue做辅助
*/
class KthLargest {
    PriorityQueue<Integer> minHeap;
    int k;
    
    //constructor
    public KthLargest (int k, int[] nums) {
        this.k = k;
        minHeap = new PriorityQueue<>(k); // Java PriorityQueue is a min heap in default
        for(int n : nums){
            add(n);
        }
    }

    public int add(int val) {
        if(minHeap.size() < k){
            minHeap.offer(val);
        }else if(minHeap.peek() < val){
            minHeap.poll();
            minHeap.offer(val);
        }
        return minHeap.peek();
    }
}