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Arranging Intergers from large to small

 
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Arranging Intergers from large to small, is it possible without an array?
overclocked
post 12 Mar, 2008 - 04:09 PM
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I was wondering if you can arrange integers without an array.(I know you can with an array) I'm trying to finish the last part of my program and I noticed my integers were not arrays so I couldn't arrange with my previous code.

Which was this
cpp
for (x=0; x<20; x++)		//In loop
    {
    cout <<"Enter your number: " << endl;
    cin >>array[x];
    }
    cout <<endl<<endl;
    cout <<"Numbers in ascending order: ";
  for (x=0; x<20; x++)
    {
    for (w=0; w<20; w++)
      {
    	if (array[w] > array[w+1])		// Sort loop
      	{
         temp = array[w];
         array[w] = array [w+1];
         array[w+1] = temp;
         }
      }
    }


I only need to arrange 5 numbers but I tried taking the same idea thats with the array but it doesn't work. Any help? I'm really not trying to rewrite the whole program... cool.gif
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KYA
post 12 Mar, 2008 - 04:25 PM
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#include <nerd.h>

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yes, but you'd have to define the sort function yourself though
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NickDMax
post 12 Mar, 2008 - 06:16 PM
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2B||!2B

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ooo neat challenge problem: Arrange an arbitrary list of numbers without the use of a typical data structure.

Well, basically you will need to use SOME structure to hold the data (even if it is just 5 variable a1, a2, a3... etc). Now there are LOTS of data structures you can choose from (linked lists, stacks, queues, trees... just to name a few). But you don't even really have to use a data structure as you can use logical program structures as well...

To arrange 5 numbers the easiest method I can think of is a little loop with an if-else-if statement...
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NickDMax
post 12 Mar, 2008 - 06:50 PM
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2B||!2B

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ok... I take that back.. the easiest method I could think of just uses a set of if-statements.
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