我的代码似乎回答了一个关于codility的测试用例的错误答案。（其他测试用例都是正确的）
测试用例： large_random
大型随机试验，N = ~100,000 我得到了一个
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
答案 0 :(得分：1)
Python 3
对于那些正在寻找易于理解的分步说明的人，这里是我的 100% 解决方案。
它通过附加和弹出来使用队列/堆栈，正如课程所建议的那样。
您可以在底部添加各种测试。提交前不要忘记删除打印输出！
def solution(A, B):
N = len(A)
print('START')
print('Fish sizes: ' + str(A))
print('Fish directions: ' + str(B))
print()
upstream = 0
downstream = []
for n in range(N):
fs = A[n]
fd = B[n]
print('Fish pos: ' + str(n) + ', fish dir: ' + str(fd)+ ', fish size: ' + str(fs))
if fd == 0 and downstream == []:
print('No fishes in the way')
upstream += 1
print('Fishes gone upstream: ' + str(upstream))
elif fd == 1:
downstream.append(fs)
print('No fishes in the way')
print('Fishes going downstream: ' + str(downstream))
elif fd == 0 and downstream != []:
print('Challenge!')
while downstream:
# pick the downstream head
pfs = downstream.pop()
# if it's bigger kills upstream, puts back downstream, move on
if pfs > fs:
downstream.append(pfs)
print('Downward eats upward')
print('Fishes going downstream: ' + str(downstream))
break
# if it's smaller don't put it back and move on to next fish
elif pfs < fs:
print('Upward eats downward')
print('Fishes going downstream: ' + str(downstream))
if downstream == []:
upstream += 1
print('Fishes gone upstream: ' + str(upstream))
break
continue
print()
print('FINISH')
print('Fishes gone upstream: ' + str(upstream))
print('Fishes going downstream: ' + str(len(downstream)))
fishes = upstream + len(downstream)
print('Fishes alive: ' + str(fishes))
return fishes
if __name__ == '__main__':
assert solution([4,3,2,1,5], [0,1,0,0,0]) == 2
print('Test 1 OK!')
print()
assert solution([1,2,3,4,5], [1,1,1,1,0]) == 1
print('Test 2 OK!')
print()
assert solution([5,4,3,2,1], [1,0,0,0,0]) == 1
print('Test 3 OK!')
print()
assert solution([1,2,3,4,5], [1,0,0,0,0]) == 4
print('Test 4 OK!')
print()
assert solution([1,2,3,4,5], [1,1,1,1,1]) == 5
print('Test 4 OK!')
print()
这是一步一步清楚解释的输出：
START
Fish sizes: [4, 3, 2, 1, 5]
Fish directions: [0, 1, 0, 0, 0]
Fish pos: 0, fish dir: 0, fish size: 4
No fishes in the way
Fishes gone upstream: 1
Fish pos: 1, fish dir: 1, fish size: 3
No fishes in the way
Fishes going downstream: [3]
Fish pos: 2, fish dir: 0, fish size: 2
Challenge!
Downward eats upward
Fishes going downstream: [3]
Fish pos: 3, fish dir: 0, fish size: 1
Challenge!
Downward eats upward
Fishes going downstream: [3]
Fish pos: 4, fish dir: 0, fish size: 5
Challenge!
Upward eats downward
Fishes going downstream: []
Fishes gone upstream: 2
FINISH
Fishes gone upstream: 2
Fishes going downstream: 0
Fishes alive: 2
Test 1 OK!
START
Fish sizes: [1, 2, 3, 4, 5]
Fish directions: [1, 1, 1, 1, 0]
Fish pos: 0, fish dir: 1, fish size: 1
No fishes in the way
Fishes going downstream: [1]
Fish pos: 1, fish dir: 1, fish size: 2
No fishes in the way
Fishes going downstream: [1, 2]
Fish pos: 2, fish dir: 1, fish size: 3
No fishes in the way
Fishes going downstream: [1, 2, 3]
Fish pos: 3, fish dir: 1, fish size: 4
No fishes in the way
Fishes going downstream: [1, 2, 3, 4]
Fish pos: 4, fish dir: 0, fish size: 5
Challenge!
Upward eats downward
Fishes going downstream: [1, 2, 3]
Upward eats downward
Fishes going downstream: [1, 2]
Upward eats downward
Fishes going downstream: [1]
Upward eats downward
Fishes going downstream: []
Fishes gone upstream: 1
FINISH
Fishes gone upstream: 1
Fishes going downstream: 0
Fishes alive: 1
Test 2 OK!
START
Fish sizes: [5, 4, 3, 2, 1]
Fish directions: [1, 0, 0, 0, 0]
Fish pos: 0, fish dir: 1, fish size: 5
No fishes in the way
Fishes going downstream: [5]
Fish pos: 1, fish dir: 0, fish size: 4
Challenge!
Downward eats upward
Fishes going downstream: [5]
Fish pos: 2, fish dir: 0, fish size: 3
Challenge!
Downward eats upward
Fishes going downstream: [5]
Fish pos: 3, fish dir: 0, fish size: 2
Challenge!
Downward eats upward
Fishes going downstream: [5]
Fish pos: 4, fish dir: 0, fish size: 1
Challenge!
Downward eats upward
Fishes going downstream: [5]
FINISH
Fishes gone upstream: 0
Fishes going downstream: 1
Fishes alive: 1
Test 3 OK!
START
Fish sizes: [1, 2, 3, 4, 5]
Fish directions: [1, 0, 0, 0, 0]
Fish pos: 0, fish dir: 1, fish size: 1
No fishes in the way
Fishes going downstream: [1]
Fish pos: 1, fish dir: 0, fish size: 2
Challenge!
Upward eats downward
Fishes going downstream: []
Fishes gone upstream: 1
Fish pos: 2, fish dir: 0, fish size: 3
No fishes in the way
Fishes gone upstream: 2
Fish pos: 3, fish dir: 0, fish size: 4
No fishes in the way
Fishes gone upstream: 3
Fish pos: 4, fish dir: 0, fish size: 5
No fishes in the way
Fishes gone upstream: 4
FINISH
Fishes gone upstream: 4
Fishes going downstream: 0
Fishes alive: 4
Test 4 OK!
START
Fish sizes: [1, 2, 3, 4, 5]
Fish directions: [1, 1, 1, 1, 1]
Fish pos: 0, fish dir: 1, fish size: 1
No fishes in the way
Fishes going downstream: [1]
Fish pos: 1, fish dir: 1, fish size: 2
No fishes in the way
Fishes going downstream: [1, 2]
Fish pos: 2, fish dir: 1, fish size: 3
No fishes in the way
Fishes going downstream: [1, 2, 3]
Fish pos: 3, fish dir: 1, fish size: 4
No fishes in the way
Fishes going downstream: [1, 2, 3, 4]
Fish pos: 4, fish dir: 1, fish size: 5
No fishes in the way
Fishes going downstream: [1, 2, 3, 4, 5]
FINISH
Fishes gone upstream: 0
Fishes going downstream: 5
Fishes alive: 5
Test 4 OK!
>
答案 1 :(得分：0)
您的整体策略对我没有意义，而且我对大多数测试用例通过感到惊讶。这是一个非常简单的例子，你的代码出错：
A:[1,2] B:[0,1]
如果我正确地理解了这个符号，鱼0是最上游的鱼，它在上游游泳，鱼1是最下游的鱼，它在下游游泳。因此，鱼不会见面，答案是两个。
但是，您的代码输出1，因为它根本没有实际检查它们的相对位置。它只是将最上游的下游鱼类与最上游的上游鱼类进行比较，并声明一个人会吃掉另一个，无论他们是否相遇。
（如果我向后收到符号，只需反转数组B - 在这两种情况下你的代码输出1，这绝对不正确。）
正确的解决方案需要比较鱼的位置，而不仅仅是它们的大小。
答案 2 :(得分：0)
如果有人仍然对此问题的解决方案感兴趣，这是我使用Python的元组方法。我得到了100％。
def solution(A, B):
fishes, stack = [], []
for i in range(len(A)):
fishes.append((B[i], A[i]))
while fishes:
if stack and not stack[-1][0] and fishes[-1][0]:
if stack[-1][1] > fishes[-1][1]:
fishes.pop()
else:
stack.pop()
else:
stack.append(fishes.pop())
return len(stack)
答案 3 :(得分：0)
这是我的方法-也许有人可以理解-我是如何解决的 Codility Python
中的Codility 100％代码
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
答案 4 :(得分：0)
我发现最优雅的Codility fish solution可以在这里找到。 它正在使用堆栈：
def solution(a, b):
l = 0 # left
s = [] # stack
for i in range(len(a)):
if b[i]==1: # fish moving upwards
s.append(a[i]) # to the stack
else: # eat fish mowing downwards.
while len(s)>0:
if s[-1]<a[i]:
s.pop() # eat from the stack
else:
break
else:
l+=1
l+=len(s)
return l