题目描述:
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.
class KthLargest {
public:
KthLargest(int k, vector<int> nums) {
K=k;
for(int i=0;i<nums.size();i++)
{
min_heap.push(nums[i]);
while(min_heap.size()>K) min_heap.pop();
}
}
int add(int val) {
min_heap.push(val);
while(min_heap.size()>K) min_heap.pop();
return min_heap.top();
}
private:
priority_queue<int, vector<int>, greater<int>> min_heap;
int K;
};
本文介绍了一个名为KthLargest的类的设计,该类能够在接收到一系列整数流时,高效地找到第K大的元素。通过使用最小堆数据结构,KthLargest能够动态维护一个大小为K的堆,确保每次调用add方法时都能返回当前流中的第K大元素。
440

被折叠的 条评论
为什么被折叠?



