arrays and index [closed]

问题

It's difficult to tell what is being asked here. This question is ambiguous, vague, incomplete, overly broad, or rhetorical and cannot be reasonably answered in its current form. For help clarifying this question so that it can be reopened, visit the help center.

Closed 9 years ago.

How can I enter numbers into an array such that duplicate entries are ignored?

For example, if I put 6 and then 3 into the array, attempting to then insert 6 into the the array should cause 6 to be rejected (since it is already in the array).

#include <iostream>

using namespace std;
int main()
{
  int x,y;
  int number;
  int arr[5];

  for (x=0; x<5; )
  {
    cout<<"enter a number:"<<endl;
    cin>>number;
    bool replace = True;
    for (y=0; y<x; y++)
    {
       if (number != arr[y])
       {
         cout << "try next time" << endl;
         replace = False;
         break;
       }
    }

    if (replace)
    {
      arr[x] = number;
      x++;
    }
  }
  return 0;
}

回答1:

std::set<int> would do what you want. This is not indexable, though.

You could use Boost.MultiIndex to give you random access and enforce uniqueness on the same underlying list of values.

btw - asking directly for code is not recommended practice.

回答2:

you have too many x++'s and you don't preset arr (maybe more style than error)

how do you know it's not working? (put some debug code inside of if (number == arr[y]) and if (replace)

回答3:

What you really want is a set. Sets cannot contain duplicate elements.

Here is a reference to the set in C++.

Just use the set as a container for your numbers. When you try to add a duplicate, it will be automatically rejected.

回答4:

You don't want an array but a datastructure called Hashtable for that;

Alternatively, you might want to look up a datastructure called associative array.

回答5:

You shouldn't use arrays for this. You should use, for example, std::set. Or, if you need to have an array as your data structure, you could encapsulate the array (e.g. realized through std::vector) in a class and define specific functions to access the array elements. Additionally, you could hold a std::set to provide a fast check for existing elements.

回答6:

Should be :

int arr[5] = {0,0,0,0,0};

Remove the x++ from the following line:

for (x=0;x<5;x++)

Then:

bool replace=true;
for (y=0;y<x;y++)
{
   if (number == arr[y])
   {
      replace=false;
      break;
   }
}

if (replace)
{
      arr[x]=number;
      x++;
}

Finally, remove the :

else if(number == arr[x])
{
    arr[x]=number;

cout << "try next time"<<endl;
}

You can insert :

cout << "try next time"<<endl;

before the

replace=false; 

回答7:

Take out the x++ in the for loop, That way you will only increment that count when you enter a new number.

Also, if you want to only run the loop five times, your outer for loop should be only to x<5.

All in all your outer loop should read:

for (x=0;x<5;)

回答8:

Take a closer look at where you increment x.

回答9:

It looks like you want to read in a sequence of numbers eliminating any duplicates. It also appears that the maximum number of unique numbers is 5.

int n = 0;  /* The number of unique numbers read in so far */
for {;;}
  cout << "enter nmber" << endl;
  cin >> number;
  for (x=0; x < n; ++x) {
    if (number == arr[x]) goto L1;  /* I love messing with peoples head by using this goto */
  }
  arr[n] = number;
  ++n;
  if (n == 5) break;
L1:
  continue;
}

来源：https://stackoverflow.com/questions/4430835/arrays-and-index