问题描述：
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 kand 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：
考虑堆的应用，heapq
代码：
1 class KthLargest: 2 3 def __init__(self, k: int, nums: List[int]): 4 nums.sort() 5 if len(nums) > k - 1: 6 nums = nums[-k:] 7 self.k = k 8 self.nums = nums 9 heapq.heapify(self.nums) 10 11 12 13 14 def add(self, val: int) -> int: 15 if len(self.nums) == self.k - 1: 16 heapq.heappush(self.nums,val) 17 elif val > self.nums[0]: 18 heapq.heapreplace(self.nums,val) 19 return self.nums[0] 20 21 22 23 24 # Your KthLargest object will be instantiated and called as such: 25 # obj = KthLargest(k, nums) 26 # param_1 = obj.add(val)
heapify(x) #以线性时间将一个列表转化为堆
heappush(heap,item) #往堆中插入一条新的值
item = heappop(heap) #从堆中弹出最小值
item = heap[0] #查看堆中最小值，不弹出
item = heapreplace(heap,item) #弹出并返回最小值，然后将heapqreplace方法中item的值插入到堆中，堆的整体结构不会发生改变