大约有 35,100 项符合查询结果（耗时：0.0291秒） [XML]
Extract part of a regex match
... answered Aug 25 '09 at 10:29
Krzysztof KrasońKrzysztof Krasoń
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Remove the last character from a string [duplicate]
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Peter Mortensen
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answered Apr 8 '11 at 9:16
anonanon
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in javascript, how can i get the last character in a string [duplicate]
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Jamie DixonJamie Dixon
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Does .NET provide an easy way convert bytes to KB, MB, GB, etc.?
...:
static readonly string[] SizeSuffixes =
{ "bytes", "KB", "MB", "GB", "TB", "PB", "EB", "ZB", "YB" };
static string SizeSuffix(Int64 value, int decimalPlaces = 1)
{
if (decimalPlaces < 0) { throw new ArgumentOutOfRangeException("decimalPlaces"); }
if (value < 0) {...
Scala best way of turning a Collection into a Map-by-key?
...on T (of type P , say), what is the best way to do a map-by-extracting-key ?
13 Answers
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Stripping everything but alphanumeric chars from a string in Python
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Otto AllmendingerOtto Allmendinger
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When should I use Kruskal as opposed to Prim (and vice versa)?
I was wondering when one should use Prim's algorithm and when Kruskal's to find the minimum spanning tree? They both have easy logics, same worst cases, and only difference is implementation which might involve a bit different data structures. So what is the deciding factor?
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How to print a dictionary line by line in Python?
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I know this is old, but I thought it would be worth mentioning that this doesn't work if cars[x] is integers. It isn't what the OP was requesting, so I'm just saying it for anybody that stumbles upon this assuming it's a blanke...
How to maintain a Unique List in Java?
...), assuming the hash function disperses the elements properly among the buckets. Iterating over this set requires time proportional to the sum of the HashSet instance's size (the number of elements) plus the "capacity" of the backing HashMap instance (the number of buckets). Thus, it's very importan...
How to split text without spaces into list of words?
...s to assume all words are independently distributed. Then you only need to know the relative frequency of all words. It is reasonable to assume that they follow Zipf's law, that is the word with rank n in the list of words has probability roughly 1/(n log N) where N is the number of words in the dic...