答案:超市排队【难度：2级】--景越训练营Python习题答案（含多种解法）

超市排队【难度：2级】:

答案1：

def queue_time(customers, n):
    l=[0]*n
    for i in customers:
        l[l.index(min(l))]+=i
    return max(l)​

答案2：

def queue_time(customers, n):
    l = [0]*n
    for i in customers:
        l[l.index(min(l))] += i
    return max(l)
    # TODO​

答案3：

def queue_time(customers, n):
    # TODO
    l=[0]*n
    for i in customers:
        l[l.index(min(l))]+=i
    return (max(l))​

答案4：

def queue_time(customers, n):
    # TODO
    l=[0]*n
    for i in customers:
        l[l.index(min(l))]+=i
    return max(l)​

答案5：

O(N^3)​

答案6：

O(N log N)​

答案7：

O(N x M^2)​

答案8：

O(n^3)​

答案9：

O(N²)​

答案10：

index(min(...))​

答案11：

O(N*(M+M)) <=> O(N*2M) <=> O(N*M) <=> O(N²)​

答案12：

import heapq

def queue_time(customers, n):
    heap = [0] * n
    for time in customers:
        heapq.heapreplace(heap, heap[0] + time)
    return max(heap)
    ​

答案13：

def queue_time(customers, n):
    qn = [0] * n
    for c in customers:
        qn = sorted(qn)
        qn[0] += c
    return max(qn)​

答案14：

O(N²*log N)​

答案15：

def queue_time(customers, n):
    queues = [0] * n
    for i in customers:
        queues.sort()
        queues[0] += i
    return max(queues)​

答案16：

class MarketQueue():
    
    def __init__(self,customers,n):
        self.customers = customers
        self.n=n
        self.timer = 0
        self.active_checkouts = []
        
    def calculate_total_time(self):
        while self.customers:
            self.process_queue()   
        return self.timer

    def process_queue(self):
        if len(self.active_checkouts) < self.n:
            queue_index = self.n - len(self.active_checkouts)
            self.active_checkouts.extend(self.customers[:queue_index])
            self.customers[:queue_index] = []
        while self.active_checkouts and (len(self.active_checkouts) == self.n or not self.customers) :
            self.timer += 1
            self.process_active_checkouts()
    
    def process_active_checkouts(self):
        finished_customers = []
        for index,customer in enumerate(self.active_checkouts):
            if customer > 1:
                self.active_checkouts[index] = int(customer-1)
            else:
                finished_customers.append(customer)
        
        for finished in finished_customers:
            self.active_checkouts.remove(finished)

# implementing requirements
def queue_time(customers,n):
    return MarketQueue(customers,n).calculate_total_time()​

答案17：

def queue_time(customers, n):
    tills = [0] * n
    for i in customers:
        indexMin = tills.index(min(tills))
        tills[indexMin] += i
    return max(tills)
​

答案18：

def queue_time(customers, n):
    tills = [0 for i in range(n)]

    for time in customers:
        min_index = tills.index(min(tills))
        tills[min_index] += time

    return max(tills)​

答案19：

def queue_time(customers, n):
    if not customers:
        return 0
    tills = customers[:n]
    customers = customers[n:]
    customers.reverse()
    while customers:
        tills.sort()
        tills[0] += customers.pop()
    return max(tills)​

答案20：

def queue_time(cu, n):
    q=[0]*n
    for i in cu:
        s=q.index(min(q))
        q[s]+=i
    return max(q)​

答案21：

def queue_time(customers, n):
    time = 0
    while len(customers[:])>0:
      time=time+1
      customers[:n]=[x-1 for x in customers[:n]]
      customers=[y for y in customers if y !=0]
    return time​