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

LeetCodeCN 第703题链接

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)

下一题：239. 滑动窗口最大值 Sliding Window Maximum