Printable View
Does any one happen to have an algorithm to pick
only the characters that appear in a String one time?
Im trying to write one but im having a hard time.
Code:for(int index = 0; index <= key.length() - 1; ++index){
char matchingchar = key.charAt(index);
for(int k = 1; k <= key.length() - (2 + index); ++k){
if(matchingchar == key.charAt(k)){
charmatch = true;
break;
}
}
if(charmatch == false){++onlyappearsonce;}
charmatch = false;
}
are you using a character array or the string header?
Actually im doing it in Java, but the Java fourm is dead as usual
Im not too sure what you mean by header. Im just passing a string to a function that is supposed to calculate hoe many characters appear only once within the string.
I came up with this and it should work. But i cant get the
if(matchingchar == key.charAt(q)) part because say for a string
"abcde" q would equal 3 then 2 then 1 so i would be compairing
matchingchar with the 3rd index which is c instead of b. not what i want. so ill have to work on this. Thanks though.
Code:
for(int index = 0; index <= key.length() - 1; ++index){
char matchingchar = key.charAt(index);
x = (key.length() * (key.length() - 2));
x =+ (key.length() - 2);
while(q != 0){
q = x % key.length();
if(matchingchar == key.charAt(q)){
++duplicates;
break;
}
--x;
}
}
int nonduplicatechars = (key.length() - (duplicates));
How long are the strings? How many types of characters are used and how are they used?
Generally I think the fastest way for not too short strings or Strings with generally few reoccurring characters, would be to set up a table of trinary values for each character, and then run trough the string once flagging on characters from 0(not used) to 1(used once) and from 1 to 2(used more than once). Then count the 1's in the table.
Hey kedaman. you seen to know C++ pretty well. How could i make this expression legal? q and x are integer types.
while(q = x % key.length() != 0){
This is somthing that i thought might work, but maybe im just going crazy thinking im on the right track.
For the first loop matchingchar = key.charAt(index);
will give me the first character then
x = (key.length() * (key.length() - 2));
x =+ (key.length() - 2);
will give me for a string say "ABCDF"
15 = 5 * 3
18 =+ 5 - 2
then
q = x % key.length() != 0
3 = 18 % 5
2 = 17 % 5
1 = 16 % 5
that would give me the correct amount of iterations to get to the end of the string each time no matter the size of the string.
I think :p
Code:for(int index = 0; index <= key.length() - 1; ++index){
matchingchar = key.charAt(index);
x = (key.length() * (key.length() - 2));
x =+ (key.length() - 2);
while(q = x % key.length() != 0){
if(matchingchar == key.charAt(z)){
++duplicates;
}
--x;
++z;
}
}
int nonduplicatechars = (key.length() - (duplicates));
while((q = x % key.length()) != 0) { .. }
Z.
Kool, ill give it a try....... Thanks
I have no idea what solution you're doing, may i even don't know what the problem is. If you fail, I could give you a hand finding the amount of duplicates
Thanks. If i do fail. Ill give you a yell. :D
I came up with this and it seems to work pretty good but the problem i am having as far as i can see is that i have to take account not only for duplicate characters but for multiple characters (3 or more) I tried this out on words like Kaaat (not that this is a word) and the results are off. And also i have to figure out how to take into account if the multiple characters are seperated. for instance. Kaata. Right now all i am doing is subtracting 2 which takes care of the original and the duplicate.
Any ideas?
Code:for(int index = 0; index <= key.length() - 2; ++index){
matchingchar = key.charAt(index);
x = (((key.length() - 2) + (key.length() * (key.length() - 2))) - index);
while((q = x % key.length()) != 0){
if(key.charAt(z) == matchingchar){
duplicates =+ 2;
}
--x;
++z;
}
z =+ 1;
}
int nonduplicatechars = (key.length() - duplicates);
screw it and go with the trinary table
What is that? Exactly. :confused:
An array of 3 different values, you could use char for instance.
each value represents a character and their value represents:
0 - Not used
1 - Used once
2 - Used twice or more
Is there any way i can mimic this in Java?
I havent coded in C++ in ages.
Of course, I know some Java, but I'm rusty on the string class, so you could make an array of char's instead. I probably will type something wrong but you'll know what I mean.
PHP Code:char ttable[256] = new char[256]
int l = key.length();
for (int x = 0; x<l ; ++x; ){
char matchingchar = key.charAt(x);
if (ttable[matchingchar]<2) ++ttable[matchingchar];
}
for (x = 0; x<256 ; ++x; ){ char count += ttable[x] == 1;}
in C and skipping the ternary aspect, simply adding.
PHP Code:void blah(char *ch)
{
int arr[256];
char tmp[256];
int j = 0;
int i = 0;
for (i = 0;i < 256;i++) { arr[i] = 0; }
for (i=0;i< (int) strlen(ch);i++) {
++arr[ ch[i] ];
}
for (i=33;i<256;i++) {
if (arr[i] == 1) { tmp[j++] = i; } //33 to ignore sdpaces
}
tmp[j] = '\0'; // make it a valid sz string
}
here again in Java
but it depends what you count as "character" so i'll leave thatPHP Code:int ttable[256] = new int[256]
int l = key.length();
for (int x = 0; x<l ; ++x; )
++ttable[key.charAt(x)];
for (x = 0; x<256 ; ++x; )
char count += ttable[x] == 1; // note no conditional stuff here either
Hey thanks guys for posting so code. Mabey you should start asking for some $$ for your help. :p Anyway..... kedaman the code that you posted, im not too sure if i understand.
for instance ++ttable[key.charAt(x)]; you would be storing the whole string within the array. correct?Code:int ttable[256] = new int[256]
int l = key.length();
for (int x = 0; x<l ; ++x; )
++ttable[key.charAt(x)];
for (x = 0; x<256 ; ++x; )
char count += ttable[x] == 1;
and this line char count += ttable[x] == 1; why would we be testing a char to see if it is equal to the length of the string?
No actually, from what I understood you want to count the amount of non duplicated characters, that is characters that are used only once. If that's not the case then you have to explain again.
++ttable[key.charAt(x)];
ttable is a table of all the characters from 0 to 255, each element specifying how many times it has been used. key.charAt(x) gets the current character and then that characters counter is incremented in the table.
char count += ttable[x] == 1;
here in this loop the count is incremented only when the current character count is 1, == returns 1 if the comparation is true otherways 0. see the logic? There's no need for conditional code
So actually each time you encounter a character you are incrementing by 1. Ok now i see. :p