說到比對就想到 regexp....

find_smallest_cycel.py

"""
Given cyclic list with elements integers.
Find the smallest cycle.
>>> find_smallest_cycle([10, 3, 999, 24, 10, 3, 999, 24, 10, 3])
[10, 3, 999, 24]
"""

def origin(mylist):
    def make_tuple(list_a):
        return list(set(map(tuple,list_a)))

    for i in range(1, len(mylist)+1):
        ans_list = [mylist[j:j+i] for j in range(0,len(mylist), i)]
        if len(ans_list) == 1:
            return ans_list[0]
        if (len(make_tuple(ans_list[:-1])) == 1
            and ans_list[0][0:len(ans_list[-1])] == ans_list[-1]):
            return ans_list[0]

def with_regexp(given_list):
    import re
    s = ' '.join(map(str, given_list))
    m = re.match(r'^((.+?)[ \d]*?)(?: \1)* \2$|^(.+?)(?: \3)*$', s)
    return list(map(int, (m.group(1) or m.group(3)).split(' ')))

in_oneline = lambda L: list(map(int,next(i for i in __import__("re").match(r'^((.+?)[ \d]*?)(?: \1)* \2$|^(.+?)(?: \3)*$',' '.join(map(str,L))).groups()if i).split(' ')))

TEST_DATA = (
    ([1], [1]),
    ([1], [1,1]),
    ([1], [1,1,1,1,1]),
    ([1,1,1,1,1,2], [1,1,1,1,1,2]),
    ([1,2], [1,2,1,2,1,2]),
    ([1,2], [1,2,1,2,1,2,1,2,1]),
    ([1,2,3], [1,2,3]),
    ([1,2,3], [1,2,3,1,2]),
    ([1,2,3], [1,2,3,1,2,3]),
    ([1,2,3], [1,2,3,1,2,3,1,2]),
    )

solutions = (
    origin,
    with_regexp,
    in_oneline,
    )

for find_smallest_cycle in solutions:
    for cycle, given_list in TEST_DATA:
        assert cycle == find_smallest_cycle(given_list)

cycle = re.compile(r'^(.*?) ?(.*? \1)*$')                                       
for x in (a, b, c):                                                             
    m = cycle.match(' '.join(map(str, x))[::-1])                                
    result = m.group(2) or m.group(1)                                           
    print([int(n) for n in result[::-1].split()])

打完收工。

改掉 find_cycle 就行了

def find_cycle(given_list):                                                     
    import re                                                                   
    s = ' '.join(map(str, given_list))[::-1]                                    
    m = re.match(r'^(.*?) ?(.*? \1)*$', s)                                      
    c = m.group(2) or m.group(1)                                                
    return list(map(int, c[::-1].split()))

m = re.match(r'^(.*?) ?(.*? \1)*$', s)
這行可以把 ^ 拿掉

額 測資如果是 [1, 1, 3, 1] 的話會錯 囧