c++ sort: ranking list with same value of an array

Question

I'm trying to show a ranking list of my array qt, which contains 5 numbers.

int i, j;
int qt[5] = {10,20,10,50,20};
int tempqt;
for (i=0; i<5; i++)
{
    for(j=(i+1); j<5; j++)
    {
        if (qt[i] >= qt[j])
        {
            tempqt = qt[i];
            qt[i] = qt[j];
            qt[j] = tempqt;
        }
    }
}

for(i=0; i<5; i++)
{
    cout << i+1 << ".number: " << qt[i] << endl;
}

normally, the 2 for-loops sort my array and the last for-loop displays my array ordered, so it looks like this:

1.number: 10
2.number: 10
3.number: 20
4.number: 20
5.number: 50

But I want to display the numbers with the same value as the same ranking position, so like this:

1.number: 10
1.number: 10
2.number: 20
2.number: 20
3.number: 50

user2556165 · Accepted Answer · 2014-11-10 21:57:28Z

2

The idea is to increase rank counter when meet different value in qt array.

i = 0;
int rank = 1, val = qt[i];
cout << rank << ".number: " << qt[i] << endl;

for(i=1; i<5; i++)
{
    if (qt[i] != val) {
        ++rank;
        val = qt[i];
    }
    cout << rank << ".number: " << qt[i] << endl;
}

answered Nov 10, 2014 at 21:57

user2556165

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Comments

PaulMcKenzie · Accepted Answer · 2014-11-10 22:41:19Z

Use std::sort to sort the array -- you won't get anywhere until the array is sorted.

#include <iostream>
#include <algorithm>

using namespace std;

int main()
{
   int qt[5] = { 10, 20, 10, 50, 20 };
   sort(qt, qt + 5);
   int count = 1;
   for (int i = 0; i < 5; ++i)
   {
       if (i > 0)
       {
          if (qt[i] != qt[i - 1])
             ++count;
       }
       cout << count << ".number: " << qt[i] << endl;
    }
}

Here is another solution using a map. This is more "lazy" in that there is no real "check if number already seen" logic involved. Just add numbers to a map, and print out the results in a loop.

If there are no memory constraints (you will need to create a map of the numbers, of course), and/or you need the array to remain stable (not sorted), then this could be an alternative.

#include <map>
#include <iostream>
#include <algorithm>

using namespace std;

int main()
{
    int qt[5] = { 10, 20, 10, 50, 20 };
    std::map<int, int> IntMap;

    // add entries to map, adding to a counter each time
    for (int i = 0; i < 5; ++i)
       IntMap[qt[i]]++;

    // output the results.
    int count = 1;
    for (auto it = IntMap.begin(); it != IntMap.end(); ++it, ++count)
    {
        for (int i = 0; i < it->second; ++i)
            cout << count << ".number: " << it->first << endl;
    }
}

The map already sorts, so that's taken care of. Then the map is set up to count the number of times each number shows up, so that's taken care of. The only thing left is to write a loop that just goes through the map and prints the information.

See it here: http://ideone.com/q08SeX

hlscalon · Accepted Answer · 2014-11-10 22:08:11Z

1

I'd rather use a do while loop:

int p = 1, x = 0;
do 
{
    cout << p << ".number: " << qt[x++] << endl;
    if (x < 5 && qt[x] != qt[x-1])
        p++;
} while (x < 5);

edited Nov 10, 2014 at 22:08

answered Nov 10, 2014 at 21:59

hlscalon

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4 Comments

Red Alert Over a year ago

this is UB when x = 4 at the start of the loop

Rudi Over a year ago

the loop is now infinite

Rudi Over a year ago

My mistake. I've used << qt[x++] << name[x++] later (to display a number and a name of a person). Didn't know that x++ actually increases the x value within the [].

Red Alert Over a year ago

@Rudi Actually, x++ increases x sometime before the next sequence point.

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