Timeline for Checking if two strings are an anagram
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Aug 7, 2014 at 12:02 | comment | added | mjolka | @EmanuelePaolini it is \$O(n)\$. Chars in .NET are 16 bits though, so you would need an array of size 65,536, which makes the dictionary a better option. | |
| Aug 7, 2014 at 11:55 | comment | added | Emanuele Paolini | The dictionary implementation seems to be \$O(n)\$ if the dictionary is implemented as an hashtable. In fact one could use a fixed array of size 256 instead of the dictionary. | |
| Aug 7, 2014 at 11:18 | comment | added | Recipe | Don't forget that this code will not check for uppercase / lowercase characters, as well as foreign characters (eg. ñ != n). You could replace the s1Sorted == s2Sorted with s1Sorted.Equals(s2Sorted, StringComparison) | |
| Aug 7, 2014 at 11:15 | comment | added | mjolka | @Pimgd thank you, I don't know what I was thinking. | |
| Aug 7, 2014 at 11:15 | history | edited | mjolka | CC BY-SA 3.0 |
I'm an idiot
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| Aug 7, 2014 at 11:14 | comment | added | Pimgd | Why bother with a for loop for comparing strings character by character? Can't you simply use a string equals function? | |
| Aug 7, 2014 at 11:08 | history | edited | mjolka | CC BY-SA 3.0 |
Add missing return
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| Aug 7, 2014 at 10:45 | history | answered | mjolka | CC BY-SA 3.0 |