I want to output the unique item from specific c++ array - c++

Question

I have created a function to count the duplicate items in array .
And everything is fine . but I want to output the unique items only , and this is my problem .
My function:
void RepeatedCounter(int n){
int i, j, temp, count= 0;
int *Numbers = new int[n];
for(i=0;i<n;i++){
cout << "Enter the number (" << i+1 << "): ";
cin >> *(Numbers+i);
}
cout << "---------------------\n";
for(i=0;i<n;i++){
temp = *(Numbers+i);
for(j=0;j<n;j++){
if(temp == *(Numbers+j)){
++count;
}
}
if(*(Numbers+i+1) != temp)
cout << *(Numbers+i) << "= " << count << endl;
count= 0;
}
delete []Numbers;
}
Main function:
int Num_Of_Digits= 0;
cout << "How many numbers: ";
cin >> Num_Of_Digits;
RepeatedCounter(Num_Of_Digits);
Example:
Inputs
1
5
3
5
1
Wrong result (current output)
1= 2
5= 2
3= 1
5= 2
1= 2
What I want
1= 2
5= 2
3= 1

Answer 1

First of: read the user data into a proper dynamic container like a vector:
std::vector<int> v;
v.reserve(100);
while (true)
{
int n;
std::cout << "Enter the number: ";
if (!(std::cin >> n)) { break; }
v.push_back(n);
}
Second, make a histogram using a map:
std::map<int, unsigned int> histogram;
for (int i : v) { ++histogram[i]; }
Now output the count:
for (auto const & p : histogram)
{
std::cout << "The number " << p->first
<< " appears " << p->second << " times.\n";
}

Answer 2

This has been done for you.
#include <iostream>
#include <algorithm>
#include <iterator>
#include <vector>
int repeated_counter(int n){
std::vector<int> vec;
std::vector<int> uniques;
int t;
for(int i=0; i!=n; ++i)
{
std::cin >> t;
vec.push_back(t);
}
std::sort(vec.begin(), vec.end());
std::unique_copy(vec.begin(), vec.end(),
std::back_inserter(uniques));
for(std::vector<int>::iterator it=uniques.begin();
it!=uniques.end();
++it)
{
std::cout << *it << "="
<< std::count(vec.begin(), vec.end(), *it) << "\n";
}
return 0;
}
please refrain from newing memory like you have done it is worse in every way to using a vector.
http://www.sgi.com/tech/stl/unique_copy.html
http://en.cppreference.com/w/cpp/container/vector

Answer 3

Try this,
for(i=0;i<n;i++){
count= 0;
temp = *(Numbers+i);
bool found = false;
for(j=0;j<n;j++){
if(temp == *(Numbers+j)){
++count;
}
}
for(j=i+1;j<n;j++) {
if(temp == *(Numbers+j)){
found = true;
}
}
if(found) continue;
if(*(Numbers+i+1) != temp)
cout << *(Numbers+i) << "= " << count << endl;
}

Answer 4

The problem is you are only checking against the next number in the list.
if(*(Numbers+i+1) != temp)
cout << *(Numbers+i) << "= " << count << endl;
What you should do is loop over the front on the list (until you get to the number you're on) and check to see if any of those numbers are the same as your current number and only print out if they aren't. You could also check the number before you started the count and not do it if the number has been done already.