这个模块(build-in)实现了一个堆的数据结构，完美的解决了Top-K问题，以后解决Top-K问题的时候，直接把这个模块拿来用就可以了
注意，默认的heap是一个小顶堆！
heapq模块提供了如下几个函数：
heapq.heappush(heap, item) 把item添加到heap中（heap是一个列表）
heapq.heappop(heap) 把堆顶元素弹出，返回的就是堆顶
heapq.heappushpop(heap, item) 先把item加入到堆中，然后再pop，比heappush()再heappop()要快得多
heapq.heapreplace(heap, item) 先pop，然后再把item加入到堆中，比heappop()再heappush()要快得多
heapq.heapify(x) 将列表x进行堆调整，默认的是小顶堆
heapq.merge(*iterables) 将多个列表合并，并进行堆调整，返回的是合并后的列表的迭代器
heapq.nlargest(n, iterable, key=None) 返回最大的n个元素（Top-K问题）
heapq.nsmallest(n, iterable, key=None) 返回最小的n个元素（Top-K问题）
import heapq
import random
# Top-K
mylist = list(random.sample(range(100), 10))
k = 3
largest = heapq.nlargest(k, mylist)
smallest = heapq.nsmallest(k, mylist)
print('original list is', mylist)
print('largest-'+str(k), ' is ', largest)
print('smallest-'+str(k), ' is ', smallest)
# heapify
print('original list is', mylist)
heapq.heapify(mylist)
print('heapify list is', mylist)
# heappush & heappop
heapq.heappush(mylist, 105)
print('pushed heap is', mylist)
heapq.heappop(mylist)
print('popped heap is', mylist)
# heappushpop & heapreplace
heapq.heappushpop(mylist, 130) # heappush -> heappop
print('heappushpop', mylist)
heapq.heapreplace(mylist, 2) # heappop -> heappush
print('heapreplace', mylist)
输出结果
original list is [18, 6, 10, 24, 48, 2, 9, 25, 16, 89]
largest-3 is [89, 48, 25]
smallest-3 is [2, 6, 9]
original list is [18, 6, 10, 24, 48, 2, 9, 25, 16, 89]
heapify list is [2, 6, 9, 16, 48, 10, 18, 25, 24, 89]
pushed heap is [2, 6, 9, 16, 48, 10, 18, 25, 24, 89, 105]
popped heap is [6, 16, 9, 24, 48, 10, 18, 25, 105, 89]
heappushpop [9, 16, 10, 24, 48, 130, 18, 25, 105, 89]
heapreplace [2, 16, 10, 24, 48, 130, 18, 25, 105, 89]
leetcode 703 python
设计一个找到数据流中第K大元素的类（class）。注意是排序后的第K大元素，不是第K个不同的元素。
你的 KthLargest 类需要一个同时接收整数 k 和整数数组nums 的构造器，它包含数据流中的初始元素。每次调用 KthLargest.add，返回当前数据流中第K大的元素。
示例:
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
说明:
你可以假设 nums 的长度≥ k-1 且k ≥ 1。
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]