Balanced brackets: Difference between revisions
m (→{{header|PureBasic}}: l) |
(→{{header|Python}}: Assumes balance must be maintained for given string.) |
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... braced -= 1 |
... braced -= 1 |
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... if braced < 0: return False |
... if braced < 0: return False |
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... return |
... return braced == 0 |
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... |
... |
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>>> for txt in ',[],[][],[[][]],][,][][,[]][[]'.split(','): |
>>> for txt in ',[],[][],[[][]],][,][][,[]][[],[[][]['.split(','): |
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... print ("%r is%s balanced" % (txt, '' if balanced(txt) else ' not')) |
... print ("%r is%s balanced" % (txt, '' if balanced(txt) else ' not')) |
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... |
... |
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'][' is not balanced |
'][' is not balanced |
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'][][' is not balanced |
'][][' is not balanced |
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'[]][[]' is not balanced |
'[]][[]' is not balanced |
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'[[][][' is not balanced</lang> |
Revision as of 13:23, 20 February 2011
You are encouraged to solve this task according to the task description, using any language you may know.
Problem: Generate a string with $N
opening brackets ("["
) and $N
closing brackets ("]"
), in some arbitrary order. Determine whether the string is balanced; that is, whether it consists entirely of pairs of opening/closing brackets (in that order), none of which mis-nest.
Examples:
(empty) OK [] OK ][ NOT OK [][] OK ][][ NOT OK [[][]] OK []][[] NOT OK
Perl 6
<lang perl6>sub balanced($s) {
my $l = 0; for $s.comb { when "]" { --$l; return False if $l < 0; } when "[" { ++$l; } } return $l == 0;
}
my $N = get; my $s = (<[ ]> xx $N).pick(*).join; say "$s {balanced($s) ?? "is" !! "is not"} well-balanced"</lang>
PureBasic
<lang PureBasic>Procedure Balanced(String$)
Protected *p.Character, cnt *p=@String$ While *p\c If *p\c='[' cnt+1 ElseIf *p\c=']' cnt-1 EndIf If cnt<0: Break: EndIf *p+SizeOf(Character) Wend If cnt=0 ProcedureReturn #True EndIf
EndProcedure</lang> Testing the procedure <lang PureBasic>Debug Balanced("") ; #true Debug Balanced("[]") ; #true Debug Balanced("][") ; #false Debug Balanced("[][]") ; #true Debug Balanced("][][") ; #false Debug Balanced("[[][]]") ; #true Debug Balanced("[]][[]") ; #false</lang>
Python
<lang python>>>> def balanced(txt): ... braced = 0 ... for ch in txt: ... if ch == '[': braced += 1 ... if ch == ']': ... braced -= 1 ... if braced < 0: return False ... return braced == 0 ... >>> for txt in ',[],[][],[[][]],][,][][,[]][[],[[][]['.split(','): ... print ("%r is%s balanced" % (txt, if balanced(txt) else ' not')) ... is balanced '[]' is balanced '[][]' is balanced '[[][]]' is balanced '][' is not balanced '][][' is not balanced '[]][[]' is not balanced '[[][][' is not balanced</lang>