检测数字序列中的重复循环(python)

Question

我想知道这是一个相当"普通"或正常的做法.并不是真的在寻找像2个班轮或任何东西那样最短的答案.我只是很快将这段代码放在一起,但我不能不觉得那里有太多的东西.此外,如果有任何库可以帮助这个,这将是非常好的.

def get_cycle(line):
    nums = line.strip().split(' ')

    # 2 main loops, for x and y
    for x in range(2, len(nums)): # (starts at 2, assuming the sequence requires at least 2 members)
        for y in range(0, x):
            # if x is already in numbers before it
            if nums[x] == nums[y]:
                seq = [nums[x]] # (re)start the sequence
                adder = 1       # (re)set the adder to 1
                ok = True       # (re)set ok to be True
                # while the sequence still matches (is ok) and
                # tail of y hasn't reached start of x
                while ok and y + adder < x:
                    if nums[x + adder] == nums[y + adder]:  # if next y and x match
                        seq.append(nums[x + adder])         # add the number to sequence
                        adder += 1                          # increase adder
                    else:
                        ok = False                          # else the sequence is broken
                # if the sequence wasn't broken and has at least 2 members
                if ok and len(seq) > 1:
                    print(' '.join(seq))    # print it out, separated by an empty space
                    return

Answer 1

我可能没有正确理解这一点,但我认为正则表达式有一个非常简单的解决方案.

(.+ .+)( \1)+

这是一个例子:

>>> regex = re.compile(r'(.+ .+)( \1)+')
>>> match = regex.search('3 0 5 5 1 5 1 6 8')
>>> match.group(0)    # entire match
'5 1 5 1'
>>> match.group(1)    # repeating portion
'5 1'
>>> match.start()     # start index of repeating portion
6

>>> match = regex.search('2 0 6 3 1 6 3 1 6 3 1')
>>> match.group(1)
'6 3 1'

以下是它的工作原理,(.+ .+)将匹配至少两个数字(尽可能多)并将结果放入捕获组1. ( \1)+将匹配一个空格,后跟捕获组1的内容,至少一次.

以及字符串的扩展说明'3 0 5 5 1 5 1 6 8':

• (.+ .+)最初会匹配整个字符串,但会因为( \1)+失败而放弃字符,这种回溯将发生,直到(.+ .+)在字符串开头不能匹配,此时正则表达式引擎将在字符串中向前移动并再次尝试
• 这将发生在捕获组从第二个5开始,它将在结束时放弃字符直到'5 1'被捕获,此时正则表达式正在查找任意数量的' 5 1'for ( \1)+,它当然会找到这个并且匹配将成功