Question

我有一个清单：

a = [2, 3, 5, 6, 6, 7, 10, 11, 13, 14, 15, 16, 16, 17, 18, 20, 21]

是否可以创建一个显示不同连续元素的最长列表的函数？

请展示如何操作

在这种情况下，答案应该是：

13, 14, 15, 16, 17, 18

Answer 1

假设您的列表已排序：

>>> from itertools import groupby
>>> z = zip(a, a[1:])
>>> tmp = [list(j) for i, j in groupby(z, key=lambda x: (x[1] - x[0]) <= 1)]
>>> max(tmp, key=len)
[(13, 14), (14, 15), (15, 16), (16, 16), (16, 17), (17, 18)]
>>> list(range(_[0][0], _[-1][-1]+1))
[13, 14, 15, 16, 17, 18]

ETA：修复最后一步;

Answer 2

最简单的做法似乎是遍历列表一次，构建您可以找到的任何序列，然后打印最长的序列。

a = [2, 3, 5, 6, 6, 7, 10, 11, 13, 14, 15, 16, 16, 17, 18, 20, 21]

seqlist = [] # List of Sequences
seq = []     # Current Sequence
last = -1

for item in a:
   # Start a new sequence if the gap from the last item is too big
   if item - last > 1:
       seqlist.append(seq)
       seq = []

   # only add item to the sequence if it's not the same as the last
   if item != last:
        seq.append(item)

   last = item

# Print longest sequence found
print max(seqlist)

Answer 3

可能有更多的pythonic方式，但我现在想不到它，所以这里有一个非常基本的解决方案：

def longestDistinctConsecutiveList ( lst ):
    lst = list( set( lst ) ) # get rid of duplicated elements
    lst.sort() # sort

    s, l = 0, 0
    for i in range( len( lst ) ):
        for j in range( i, len( lst ) ):
            if lst[j] - lst[i] == len( lst[i:j] ) > l:
                l = 1 + a[j] - a[i]
                s = i
    return lst[s:s+l]

Python中最长的不同连续列表

3 个答案: