试着想象一下你在超市买了东西然后排队结账，给定一个数字列表nums和数字n，其中nums中每个数字依次代表等待结账的用户所需分钟数，n表示结账卡口个数。用户优先选择空闲卡口排队结账。 请编写一个函数，计算出需要多久可以让全部用户结账完毕。
【示例】
输入：[10,2,3,3], 2
输出：10
解释：耗时10分钟的用户进入卡口一，卡口持续占用中；耗时2分钟的用户进入卡口二，2分钟后结账完毕，此时耗时3分钟的用户进入空闲的卡口二。以此类推，全部结账完需要10分钟。
题目难度：中等
题目来源：codewars-The Supermarket Queue
def solution(nums: list, n: int)-> int:
# your code here
assert solution([5,3,4], 1) == 12
assert solution([10,2,3,3], 2) == 10
assert solution([2,3,10], 2) == 12
def solution(team,n):
a = 0
b = 0
if n==1:
for i in range(len(team)):
a=a+team[i]
return a
else:
a = team[0]
b = team[n-1]
for i in range(2,len(team)):
if a-b>0:
b=b+team[i]
else:
a=a+team[i]
if a>b:
return a
else:
return b
assert solution([5,3,4], 1) == 12
assert solution([10,2,3,3], 2) == 10
assert solution([2,3,10], 2) == 12
def solution(nums: list, n: int)-> int:
nlist = []
#初始化排队口
for i in range(n):
nlist.append(0)
for num in nums:
nlist[nlist.index(min(nlist))] += num
return max(nlist)
assert solution([10,2,3,3,12], 2) == 20 这个断言过不去
def solution(nums: list, n: int) -> int:
# your code here
queue_list = [0] * n
if n == 1:
return sum(nums)
else:
for i in nums:
queue_list[queue_list.index(min(queue_list))] += i
return max(queue_list)
assert solution([5, 3, 4], 1) == 12
assert solution([10, 2, 3, 3], 2) == 10
assert solution([2, 3, 10], 2) == 12
#果然想复杂了呢哎
def solution(nums: list, n: int)-> int:
# your code here
ent = nums[:n]
T = 0
for i in nums[n:]:
minnest = min(ent)
T += minnest
for index,value in enumerate(ent):
ent[index] = value - minnest
if ent[index] == 0:
ent[index] += i
return T + max(ent)
assert solution([5,3,4], 1) == 12
assert solution([10,2,3,3], 2) == 10
assert solution([2,3,10], 2) == 12
assert solution([10,2,3,3,12], 2) == 20
def solution(nums:list,n:int):
li=[0 for _ in range(n)]
for num in nums:
li[li.index(min(li))]+=num
return max(li)
def solution(nums: list, n: int)-> int:
if len(nums)==0:return 0
if 0<len(nums)<=n:
return max(nums)
else:
queue = [nums[i] for i in range(n)]
queue.sort(reverse=True)
for i in range(n,len(nums)):
queue[n-1] +=nums[i]
#排序
tmp=queue[n-1]
left=n-2
while left>=0 and queue[left]<tmp:
queue[left + 1] = queue[left]
left -= 1
queue[left + 1] = tmp
return queue[0]
def solution(nums: list, n: int) -> int:
# 创建卡口
lis = [0 for i in range(n)]
# 时间计算
for i in nums:
lis.sort()
lis[0] += i
print(lis)
return max(lis)
assert solution([5, 3, 4], 1) == 12
assert solution([10, 2, 3, 3], 2) == 10
assert solution([2, 3, 10], 2) == 12
def solution(nums: list, n: int)-> int:
s = 0
i = n
while n < len(nums):
nums_i = nums[0:i]
nums[0:i] = [j - min(nums_i) for j in nums_i]
nums[nums_i.index(min(nums_i))] = nums[n]
s += min(nums_i)
n += 1
return s + max(nums)
assert solution([5,3,4], 1) == 12
assert solution([10,2,3,3], 2) == 10
assert solution([2,3,10], 2) == 12
assert solution([3,2,3,6,8,10], 3) == 13def solution(nums: list, n: int)-> int:
# your code here
l = [0] * n
for i in nums:
l[l.index(min(l))] += i
return max(l)
assert solution([5,3,4], 1) == 12
assert solution([10,2,3,3], 2) == 10
assert solution([2,3,10], 2) == 12
print(solution([10,2,3,3,3,2,8,9], 5))
def solution(nums: list, n: int) -> int:
res_list = [0 for _ in range(n)]
for num in nums:
res_list[res_list.index(min(res_list))] += num
return max(res_list)
assert solution([5, 3, 4], 1) == 12
assert solution([10, 2, 3, 3], 2) == 10
assert solution([2, 3, 10], 2) == 12