有关栈、堆、队列的LeetCode做题笔记，Python实现
703. 数据流中的第K大元素 Kth Largest Element in a Stream
方法一：直接降序排序，然后取第k个元素返回，add时每次都再排序一次，这样时间复杂度为O(k*logk)
# 1.直接排序
class KthLargest:
def __init__(self, k: int, nums: List[int]):
self.nums = nums
self.k = k
self.nums.sort(reverse = True)
while len(self.nums) > k:
self.nums.pop()
def add(self, val: int) -> int:
self.nums.append(val)
self.nums.sort(reverse = True)
if len(self.nums) > self.k:
self.nums.pop()
return self.nums[-1]
方法二：使用小顶堆实现的优先队列，Python 中标准库 heapq 就是小顶堆，时间复杂度降低为O(k)
# 2.小顶堆
import heapq
class KthLargest:
def __init__(self, k: int, nums: List[int]):
self.pool = nums
heapq.heapify(self.pool)
self.k = k
while len(self.pool) > k:
heapq.heappop(self.pool)
def add(self, val: int) -> int:
if len(self.pool) < self.k:
heapq.heappush(self.pool, val)
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)
