我的代码似乎针对一个有关Codility的测试用例返回了错误的答案（其余测试用例是正确的）
对于一个测试用例：
large_random
大型随机检验，N =〜100,000
我正在
得到868预期840
问题链接：
https://app.codility.com/programmers/lessons/7-stacks_and_queues/fish/
def solution(A, B):
#declare stacks for fish traveling downstrea, upstream
dstrm = []
ustrm = []
#set counter to zero for fish traveling upstream UNHINDERED
#by fish traveling downstream, needed because iteration starts upstream
counter = 0
n = len(A)
#iterate over input, starting from upstream
for i in range(n):
if B[i] == 0:
ustrm.append(A[i])
elif B[i] == 1:
dstrm.append(A[i])
# clear upstream stack of fish known to be UNHINDERED, increase counter
if len(ustrm) > 0 and len(dstrm) == 0:
counter += len(ustrm)
ustrm.clear()
#compare what's bigger and decrease stack of fish that is eaten
while len(dstrm) > 0 and len(ustrm) > 0:
if dstrm[-1] > ustrm[-1]:
ustrm.pop()
elif ustrm[-1] > dstrm[-1]:
dstrm.pop()
return len(dstrm) + len(ustrm) + counter
最佳答案
这是我的方法-可能有人可以想到-我是如何解决的
Codility Python中的Codility 100％
码
def solution(A, B):
"""
Codility 100%
https://app.codility.com/demo/results/trainingF5HTCA-YFV/
Idea is to use stack concept
For each iteration if current fish is of type 1 add it to stack 1 fish
Create stack 1 fish - it always holds the downstream ie 1 kind of fish
and keep killing the smaller fish from the list if it is greater
else killed by current fish.
Now if stack 1 fish has no fish it means current fish can be counted as remaining fish.
0 represents a fish flowing upstream - 0 fish
1 represents a fish flowing downstream - 1 fish
A[0] = 4 B[0] = 0 alive fish
A[1] = 3 B[1] = 1 downstream
A[2] = 2 B[2] = 0 eaten by A[1]
A[3] = 1 B[3] = 0 eaten by A[1]
A[4] = 5 B[4] = 0 eat to A[1] and remain alive
"""
count = 0
# stores the 1 fish in stack
stack_1_fish = []
print(A)
print(B)
for index in range(len(A)):
if B[index] == 0:
# until stack has some 1 fish
while stack_1_fish:
# get last fish from stack and check if it can die or not
# the larger fish eats the smaller one.
# two fish P and Q meet each other when P < Q, B[P] = 1 and B[Q] = 0, and there are no living fish between them
if stack_1_fish[-1] > A[index]:
# stack 1 fish kill to current fish
# exit from while loop and else block also execute next top for loop
# check for other fish fight
print("Killed by " + str(stack_1_fish[-1]) + " Die " + str(A[index]))
break
else:
# stack 1 fish is killed by current fish
p = stack_1_fish.pop()
print("Killed by " + str(A[index]) + " Die " + str(p))
# this is the case when stack becomes empty ie. no fish of kind 1
# it would never meet previous fish but can fight with downstream fish
else:
print("Count fish as remaining......." + str(A[index]))
count += 1
else:
# fish 1 found add to stack
stack_1_fish.append(A[index])
print("1 fish found, add to stack, it can kill or killed by someone..." + str(A[index]))
print(stack_1_fish)
print("Count: " + str(count))
print("Stack 1 fish: " + str(len(stack_1_fish)))
return count + len(stack_1_fish)
执行输出-
if __name__ == '__main__':
result = solution([4, 3, 2, 1, 5], [0, 1, 0, 0, 0])
# result = solution([4, 3, 2, 1, 5], [0, 0, 0, 0, 0])
# result = solution([4, 3, 2, 1, 5], [1, 1, 1, 1, 1])
print("")
print("Solution " + str(result))
[4, 3, 2, 1, 5]
[0, 1, 0, 0, 0]
Count fish as remaining.......4
1 fish found, add to stack, it can kill or killed by someone...3
[3]
Killed by 3 Die 2
Killed by 3 Die 1
Killed by 5 Die 3
Count fish as remaining.......5
Count: 2
Stack 1 fish: 0
Solution 2
[4, 3, 2, 1, 5]
[0, 0, 0, 0, 0]
Count fish as remaining.......4
Count fish as remaining.......3
Count fish as remaining.......2
Count fish as remaining.......1
Count fish as remaining.......5
Count: 5
Stack 1 fish: 0
Solution 5
[4, 3, 2, 1, 5]
[1, 1, 1, 1, 1]
1 fish found, add to stack, it can kill or killed by someone...4
[4]
1 fish found, add to stack, it can kill or killed by someone...3
[4, 3]
1 fish found, add to stack, it can kill or killed by someone...2
[4, 3, 2]
1 fish found, add to stack, it can kill or killed by someone...1
[4, 3, 2, 1]
1 fish found, add to stack, it can kill or killed by someone...5
[4, 3, 2, 1, 5]
Count: 0
Stack 1 fish: 5
Solution 5