python - 基于公共(public)头或尾连接列表

Question

解决以下问题最快的方法是什么我将根据共同的头部或尾部加入多个列表

input = ([5,6,7], [1,2,3], [3,4,5], [8, 9])
output = [1, 2, 3, 4, 5, 6, 7]

Answer 1

>>> def chain(inp):
    d = {}
    for i in inp:
        d[i[0]] = i[:], i[-1]
    l, n = d.pop(min(d))
    while True:
        lt, n = d.pop(n, [None, None])
        if n is None:
            if len(d) == len(inp) - 1:
                l, n = d.pop(min(d))
                continue
            break
        l += lt[1:]
    return l

>>> chain(input)
[1, 2, 3, 4, 5, 6, 7]
>>> chain(([5,6,7], [1,2,10], [3,4,5], [8, 9]))
[3, 4, 5, 6, 7]