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Answer 1

+1 A:

>>> 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]

SilentGhost 2010-02-19 13:20:31

Thanks SilentGhost. It's general enough (str and int) and works perfectly. Are there any general name for this type of method. I call it 'tiling list'

Joey 2010-02-19 13:50:28

may be http://en.wikipedia.org/wiki/Linked_list ?

SilentGhost 2010-02-19 14:02:59

input = ([5,6,7], [1,2,3], [3,4,5], [8, 9], [9, 10, 11], [12])output = ([1, 2, 3, 4, 5, 6, 7], [8, 9, 10, 11])is recursion a good idea for this case?

Joey 2010-02-19 14:14:09

input = ([5,6,7], [1,2,10], [3,4,5], [8, 9])output = [3, 4, 5, 6, 7]Oeps! the previous won't work if given the above input

Joey 2010-02-19 14:31:33

given the above input, you just need to check `d` and run my code again, therefore obtaining `[8,9,10,11]` list. in your second example you could do the same, but it's not clear on what ground would you have excluded `[1, 2, 10]` and `[8,9]` lists.

SilentGhost 2010-02-19 14:34:51

[3,4,5] and [5,6,7] are joining because [3,4,5] has a tail 5 and [5,6,7] have a head 5.

Joey 2010-02-19 14:45:20

@Joey: updated the code

SilentGhost 2010-02-19 15:08:42

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join lists based on common head or tail

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