First clear solution for Letter Queue by xiongbiao - python coding challenges

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First solution in Clear category for Letter Queue by xiongbiao

def letter_queue(commands):
    l = []
    for i in commands:
        if i[1] == 'U':
            l.append(i[5:])
        else:
            if len(l) != 0:
                del l[0]
    return "".join(l)

July 13, 2014

Comments:

Cjkjvfnby on Aug. 13, 2014, 10:23 a.m.
<a href="http://www.checkio.org/forum/post/1618/proper-way-to-check-if-sequence-is-empty/">http://www.checkio.org/forum/post/1618/proper-way-to-check-if-sequence-is-empty/</a> you can use str.startswith it is more readable.

cellsuicide on Aug. 21, 2014, 9:17 p.m.
Not really clear...if you're judging based on the understanding a novice programmer would have.

xtofl on Sept. 28, 2014, 3:30 p.m.
Interesting that you would pick the second char of the command to base your decision upon. What happens when you introduce a "CUT <index>" command...? You can omit the `len` test: <pre class='brush: python'>if l: del l[0] </pre>

xiongbiao on Oct. 23, 2014, 4:36 p.m.
get it, thanks!!

texom512 on Aug. 1, 2015, 2:08 p.m.
"if i[1] == 'U':" is not clear. That's better: "if i.startswith('PUSH'):"

Strings Theory

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