703 Kth Largest Element in a Stream 数据流中的第K大元素
Description:
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.
题目描述:
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。
你的 KthLargest 类需要一个同时接收整数 k 和整数数组nums 的构造器，它包含数据流中的初始元素。每次调用 KthLargest.add，返回当前数据流中第K大的元素。
示例 :
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
说明:
你可以假设 nums 的长度≥ k-1 且k ≥ 1。
思路:
由于需要输出第 k大的元素, 最好的方案是维护一个小顶堆, 可以用优先队列实现
时间复杂度O(n), 空间复杂度O(k), n为数组 arr的大小
代码:
C++:
class KthLargest
{
public:
KthLargest(int k, vector& nums)
{
pq_size = k;
for (int i = 0; i < nums.size(); i++)
{
pq.push(nums[i]);
if (pq.size() > pq_size) pq.pop();
}
}
int add(int val)
{
pq.push(val);
if (pq.size() > pq_size) pq.pop();
return pq.top();
}
private:
priority_queue, greater> pq;
int pq_size;
};
/**
* Your KthLargest object will be instantiated and called as such:
* KthLargest* obj = new KthLargest(k, nums);
* int param_1 = obj->add(val);
*/
Java:
class KthLargest {
private PriorityQueue pq;
private int pqSize;
public KthLargest(int k, int[] nums) {
pqSize = k;
pq = new PriorityQueue(pqSize);
for (int i : nums) add(i);
}
public int add(int val) {
if (pq.size() < pqSize) pq.offer(val);
else if (pq.peek() < val) {
pq.poll();
pq.offer(val);
}
return pq.peek();
}
}
/**
* Your KthLargest object will be instantiated and called as such:
* KthLargest obj = new KthLargest(k, nums);
* int param_1 = obj.add(val);
*/
Python:
import heapq
class KthLargest:
def __init__(self, k: int, nums: List[int]):
self.k, self.pq = k, nums
heapq.heapify(self.pq)
while len(self.pq) > k:
heapq.heappop(self.pq)
def add(self, val: int) -> int:
if len(self.pq) < self.k:
heapq.heappush(self.pq, val)
elif self.pq[0] < val:
heapq.heapreplace(self.pq, val)
return self.pq[0]
# Your KthLargest object will be instantiated and called as such:
# obj = KthLargest(k, nums)
# param_1 = obj.add(val)