题目地址：https://leetcode.com/problems/kth-largest-element-in-a-stream/description/

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:

Youmay assume that nums' length ≥ k-1 and k ≥ 1.

题目大意

实现一个类，这个类能找出一个数据流中第K大的数。

解题方法

小根堆

曾经在亚马逊遇到过的面试题，可惜当时不会，甚至连用堆来实现都没想到。现在知道用堆来实现了。

Python的堆是小根堆，不需要对其进行转换，我们想一想，如果一个堆的大小是k的话，那么最小的数字就在其最前面（即为第k大的数字），只要维护当新来的数字和最前面的这个数字比较即可。

所以我们的工作就是维护一个小根堆，这个小根堆保存的是从第K大的数字到最大的数字。堆的大小即为K。

实现过程比较简单，只要熟悉Python中堆的操作即可。

代码如下：

class KthLargest(object):

    def __init__(self, k, nums):
        """
        :type k: int
        :type nums: List[int]
        """
        self.pool = nums
        self.size = len(self.pool)
        self.k = k
        heapq.heapify(self.pool)
        while self.size > k:
            heapq.heappop(self.pool)
            self.size -= 1

    def add(self, val):
        """
        :type val: int
        :rtype: int
        """
        if self.size < self.k:
            heapq.heappush(self.pool, val)
            self.size += 1
        elif val > self.pool[0]:
            heapq.heapreplace(self.pool, val)
        return self.pool[0]
# Your KthLargest object will be instantiated and called as such:
# obj = KthLargest(k, nums)
# param_1 = obj.add(val)

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