[99클럽 코테 스터디 10일차 TIL] 힙

문제

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) Appends the integer val to the stream and 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.

답

import java.util.PriorityQueue;

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

    public KthLargest(int k, int[] nums) {
        this.k = k;
        this.minHeap = new PriorityQueue<>(k);

        for (int num : nums) {
            add(num);
        }
    }

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

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

회고

오늘 문제는 이해가 잘 안가서 정확한 요구사항을 파악하지 못해 멋대로 풀다가 나중에 지피티의 도움을 받았다. 그리고 문제 카테고리가 힙인걸 보고 힙에 대해 바로 검색해서 공부하였다.
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