I have to implements a function that takes a string as an input and finds the non-duplicate character from this string.
So an an example is if I pass string str = \"
Revised version of the first answer i.e: You don't need ToCharArray() function for this to work.
public static string RemoveDuplicates(string input)
{
return new string(input.Distinct().ToArray());
}
My answer in java language.
Posting here so that you might get a idea even it is in Java language.Algorithm would remain same.
public String removeDup(String s)
{
if(s==null) return null;
int l = s.length();
//if length is less than 2 return string
if(l<2)return s;
char arr[] = s.toCharArray();
for(int i=0;i<l;i++)
{
int j =i+1; //index to check with ith index
int t = i+1; //index of first repetative char.
while(j<l)
{
if(arr[j]==arr[i])
{
j++;
}
else
{
arr[t]=arr[j];
t++;
j++;
}
}
l=t;
}
return new String(arr,0,l);
}
A Linq approach:
public static string RemoveDuplicates(string input)
{
return new string(input.ToCharArray().Distinct().ToArray());
}
Below is the code to remove duplicate chars from a string
var input = "SaaSingeshe";
var filteredString = new StringBuilder();
foreach(char c in input)
{
if(filteredString.ToString().IndexOf(c)==-1)
{
filteredString.Append(c);
}
}
Console.WriteLine(filteredString);
Console.ReadKey();
For arbitrary length strings of byte-sized characters (not for wide characters or other encodings), I would use a lookup table, one bit per character (32 bytes for a 256-bit table). Loop through your string, only output characters that don't have their bits turned on, then turn the bit on for that character.
string removedupes(string s)
{
string t;
byte[] found = new byte[256];
foreach(char c in s)
{
if(!found[c]) {
t.Append(c);
found[c]=1;
}
}
return t;
}
I am not good with C#, so I don't know the right way to use a bitfield instead of a byte array.
If you know that your strings are going to be very short, then other approaches would offer better memory usage and/or speed.
It sounds like homework to me, so I'm just going to describe at a high level.