leetcode 703 数字流中第k大的数字

题目：
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.
思路：
这个题目很简单，只需要考虑最大的k个数字就行，把这个k个数字用一个“窗口”框起来，窗口大小是k，每次新添加的数字只要比窗口的大，就让这个最小数被替代掉即可
代码：
Python
class KthLargest(object):

def __init__(self, k, nums):
    """
    :type k: int
    :type nums: List[int]
    """
    self.pool=nums
    self.k=k
    heapq.heapify(nums)
    while len(self.pool)>k:
        heapq.heappop(self.pool)

def add(self, val):
    """
    :type val: int
    :rtype: int
    """
    if len(self.pool)self.pool[0]:
        heapq.heapreplace(self.pool,val)
    return self.pool[0]

C++
class KthLargest {
private:
int K;
priority_queue pq;
public:
KthLargest(int k, vector nums) {
K = k;
for (auto n : nums)
add(n);
}

int add(int val) {
    if (pq.size() < K || val > pq.top())
        pq.push(val);
    
    if (pq.size() > K)
        pq.pop();
    
    return pq.top();
}

};