703. Kth Largest Element in a Stream

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Problem

Design a class to find the $k^{th}$ largest element in a stream. Note that it is the $k^{th}$ largest element in the sorted order, not the $k^{th}$ 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 $k^{th}$ 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:

• $1 <= k <= 10^4$
• $0 <= nums.length <= 10^4$
• $-10^4 <= nums[i] <= 10^4$
• $-10^4 <= val <= 10^4$
• At most $10^4$ 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 $k^{th}$ element.

Code

215. Kth Largest Element in an Array

注意题意

2,3,4,5,5,8

8 - 第1大
5 - 第2大
5 - 第3大
4 - 第4大

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

    public KthLargest(int k, int[] nums) {
        this.k = k;
        // 维持k的大小 也就保证了queue中存放了前k个最大值
        queue = new PriorityQueue<>();

        for (int num : nums) {
            queue.offer(num);
            if (queue.size() > k) {
                queue.poll();
            }
        }
    }

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

        return queue.peek();
    }
}

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