import heapq
def getMaxUnit(num,boxes,UnitSize,UnitSize,unitPerBox, truckSize):
if truckSize == 0 or num == 0:
return 0
h = []
for i in range(num):
h.append((-1*unitPerBox[i],boxes[i]))
heapq.heapify(h)
maxCapacity = 0
while truckSize>=0 and len(h) != 0:
popped = heapq.heappop(h)
truckSize = truckSize-popped[1]
available = popped[1]
if truckSize < 0:
available = popped[1]+truckSize
maxCapacity = maxCapacity + available*(-1*popped[0])
return maxCapacity
我试图在这里找到代码的时间复杂度。我对 heapq.heappop 的时间复杂度感到困惑,因为每次弹出元素时它都需要维护堆属性。