python中有堆吗

"ki<=k2i,ki<=k2i+1;或ki>=k2i,ki>=k2i+1.（i=1,2,…,[n/2])"

import heapq1.heapq.heappush(heap,item)  #heap为定义堆，item 增加的元素;eg.  heap=[]  heapq.heappush(heap, 2)2.heapq.heapify(list)        #将列表转换为堆eg.   list=[5,8,0,3,6,7,9,1,4,2]   heapq.heapify(list) 3.heapq.heappop(heap)        #删除最小的值eg.  heap=[2, 4, 3, 5, 7, 8, 9, 6]  heapq.heappop(heap) ---->heap=[3, 4, 5, 7, 9, 6, 8]4.heapq.heapreplace(heap, item)     #删除最小元素值，添加新的元素值eg.  heap=[2, 4, 3, 5, 7, 8, 9, 6]  heapq.heapreplace(heap, 11) ------>heap=[2, 3, 4, 6, 8, 5, 7, 9, 11]5.heapq.heappushpop(heap, item)     #首判断添加元素值与堆的第一个元素值对比，如果大于则删除最小元素，然后添加新的元素值，否则不更改堆eg.   条件：item >heap[0]   heap=[2, 4, 3, 5, 7, 8, 9, 6]   heapq.heappushpop(heap, 9)---->heap=[3, 4, 5, 6, 8, 9, 9, 7]   条件：item   heap=[2, 4, 3, 5, 7, 8, 9, 6]   heapq.heappushpop(heap, 9)---->heap=[2, 4, 3, 5, 7, 8, 9, 6]6.heapq.merge(...)             #将多个堆合并7.heapq.nlargest (n, heap)     #查询堆中的最大元素，n表示查询元素个数eg.  heap=[2, 3, 5, 6, 4, 8, 7, 9]  heapq.nlargest (1, heap)--->[9]8.heapq.nsmallest(n, heap)     #查询堆中的最小元素，n表示查询元素个数eg. heap=[2, 3, 5, 6, 4, 8, 7, 9] heapq.nlargest (1, heap)--->[2]