0

在输入-1之前，我将如何读取整数，然后打印最长的连续数字序列的长度，其中奇数和偶数交替出现？

我已经完成了第一部分，但从那里开始走下坡路。

一些测试清单：

[1,2,3,4,5,10,6,7,8,20,25,30,40,-1]              
[6,7,8,20,25,30,40,1,2,3,4,5,10,15,20,-1]

这是我的代码： 

evenOdd=[]

while True:
    try:
        n=int(input())
        if n != -1:
            evenOdd.append(n)
    except:
            break
evenOdd=[]
longest = 0
length = 0
for i in range(len(evenOdd)):
    if ((evenOdd[i-2]% 2 == 0) and (evenOdd[i-1]% 2 == 1) and (evenOdd[i]% 2 == 0):
        length += 1
    else:
        longest = max(longest, length)
        length = 0

print(longest)
4

4 回答 4

0

我就是这样做的。我认为这可能比上面的例子更简单。

def 交替（lst）：

longSeries = []
currentSeries=[]
for i in range (len(lst)-1):
    if i == 0:
        currentSeries = [lst[0]]
    if(abs(lst[i] - lst[i+1]) % 2 ==  1):
        currentSeries.append(lst[i+1])
    else:
        currentSeries = [lst[i+1]]
    if(len(currentSeries) > len(longSeries)):
        longSeries = currentSeries


print ("The longest series is: " +str(longSeries))
print(len(longSeries))
于 2018-11-26T01:32:36.017 回答
0

您可以用于itertools.cycle在余数 0 和 1 之间交替，并用于itertools.groupby对奇偶序列进行分组：

from itertools import groupby, cycle
l = [1,2,3,4,5,10,6,7,8,20,25,30,40]
r = cycle((0, 1))
print(max(sum(1 for i in g) for _, g in groupby(l, key=lambda n: n % 2 == next(r))))

这输出：（6因为最长的奇偶序列是1,2,3,4,5,10）

于 2018-10-15T16:02:35.467 回答
0

一种选择是随时跟踪最长的序列：

longest = []
current = []

while True:
    n = int(input("Enter value: "))
    if n == -1:
        break

    if current and current[-1] % 2 != n % 2:
        current.append(n)
    else:
        current = [n]

    if len(current) > len(longest):
        longest = current

这里的好处-1是输入时无需进行后处理，结果就可以使用了。

于 2018-08-08T02:42:46.677 回答
-1

您可以申请itertools.groupby两次：

import itertools
d = [[1,2,3,4,5,10,6,7,8,20,25,30,40,-1], [6,7,8,20,25,30,40,1,2,3,4,5,10,15,20,-1]]
def key_func(d):
  start= not d[0]%2
  for i in d[1:]:
    if i%2 == start:
      start = (not i%2)
    else:
      return False
  return True

for l in d:
  new_l = [list(b) for _, b in itertools.groupby(l, key=lambda x:x%2)]
  second_l = [[i for [i] in b] for a, b in itertools.groupby(new_l, key=lambda x:len(x) ==1) if a]
  print(max(second_l, key=lambda x:[key_func(x), len(x)]))

输出：

[1, 2, 3, 4, 5]
[1, 2, 3, 4, 5, 10, 15, 20, -1]
于 2018-08-08T02:39:05.220 回答