how to search properly the vector for a value - c++

Question

The problem I have, I have to add to a vector, the missing chars.
For example I have initially
s,a,p,i,e,n,t,i,a
and I have to add missing chars to it
s,a,p,i,e,n,t,i,a,b,c,d ...
I am trying to use this code to search for an existing value.
for(char c='a';c!='z';++c)
{
if (vec.end()!=find(vec.begin(),vec.end(),c))
vec.push_back(c);
}
The find returns last when it fails to locate a value. But how do I know if last value was in it?
EDIT
When the for loop starts, for 'a' returns vec.end() so it should not go in, but goes in, and adds 'a' again in the end.
See this in debugger
alt text http://img203.imageshack.us/img203/2048/bb1f.jpg
(The bug I have, the value in last position gets inserted twice, I have to omit this)

Answer 1

In your case it's best to:
Create one vector(bool), with indexes from 'a' to 'z' ,initialize it to false, (i)
Run once through your original vector, set true in the other vector the element for each character that you find,
Run once through this new vector and for each false value, append the corresponding char to the original.
(i) You may use actual_index = character - 'a'. Put some assertions here and there so that you don't crash for characters outside the range you are checking, presumably 'a' to 'z' (which by the way is not a strict definition of what a char is).
With just one initialization, two steps linear steps and no searches, you'll be done.

Answer 2

What others have answered is true but you should also change the termination condition in your for loop to c <= 'z' if you want the letter z to be included in your list.
EDIT
I can't help adding that with the Boost.RangeEx library your problem can be solved with a one-liner:
boost::set_difference(boost::counting_range('a', char('z' + 1)),
std::set<char>(vec.begin(), vec.end()),
std::back_inserter(vec));

Answer 3

Nope, end() is not the last element of the vector but past it. To iterate over all elements you normally do
for(it= vec.begin(); it!= vec.end(); it++) ...
So whatever your problem is, this is ok.

Answer 4

When find succeeds it returns iterator which is pointing to the found position. So if the value is in the vector then the return value will be something other that vec.end(). The condition inside the if condition should be == and not != if you are trying to create vector of unique characters.

Answer 5

If you need to find a value in your container, then the greatest likelihood is that you need to use a different sort of container, where searching is fast!
Have a Look at this very useful diagram by Adrinael:
(source: adrinael.net)
(In your case I believe std::set is probably the most appropriate)

Answer 6

vec.end() returns an iterator whose position is past the last element in the vector. If it matches on the last value, the returned iterator will not be equal to vec.end().

Answer 7

You want to insert the character into array if it is NOT found, when it == end.
You are checking if it != end so you insert characters when they are already found in the string. See this:
char v[] = "acef";
vector<char> str(v,v+sizeof(v));
copy(str.begin(), str.end(), ostream_iterator<char>(cout, ","));
cout << endl;
for (char c = 'a'; c < 'g'; ++c)
{
vector<char>::iterator it = find(str.begin(),str.end(), c);
if (it == str.end())
{
str.push_back(c);
}
}
copy(str.begin(), str.end(), ostream_iterator<char>(cout, ","));
output:
a,c,e,f,,
a,c,e,f,,b,d,
The extra empty character ,, is the null in the original string "acef" - null terminated.

Answer 8

you can start by sorting the vector
in this way you will notice instantly for gaps in the sequence
#include <stdio.h>
#include <stdlib.h>
#include <vector>
#include <iostream>
#include <algorithm>
using namespace std;
int main(int argc, char** argv)
{
vector<char> original;
original.push_back('a');
original.push_back('d');
original.push_back('x');
original.push_back('j');
original.push_back('z');
sort(original.begin(), original.end());
vector<char> unseen_chars;
char current_char = 0;
char last_char = original[0];
for (int i = 1; i <= original.size(); i++)
{
current_char = original[i];
for ( char j = last_char + 1; j < current_char; j++)
{
unseen_chars.push_back(j);
}
last_char = current_char;
}
for (int i = 0; i < unseen_chars.size(); i++)
{
cout << unseen_chars[i];
}
cout << endl;