检查NSArray中每对唯一的对象

Question

I have an array of numbers. I need to compare every number in the array with every other number in the array without any duplication of comparison sets. Example, need to compare objects at index 0 and 1 but don't want to double up later by checking objects at index 1 and 0.

Can anyone help me out with the algorithm for this. It would be greatly appreciated.

Answer 1

I don't know what you need to do this for as there are probably better ways to accomplish whatever you are trying to do, but for the case you are talking about, you could do a simple:

for (int n=0;n<[array count];n++) {
    for (int m=n+1;m<[array count];m++) {
        //check your array based on objects at index n and m;
    }
}

This just loops through the array from beginning to end, and for each object loops through every object after it and you can compare them or do whatever. Starting the inner loop at n+1 instead of 0 prevents you from repeating comparisons.

Answer 2

Use an NSMutableOrderedSet if it helps solve your problem. Or another mutable array and add objects from your original array to it one by one using containsObject: to test if the object youre about to insert will be a duplicate.

Answer 3

You can iterate over the array and at each number compare it with the next numbers in the array only. if you have array with size 5, then at arr[2] compare it with arr[3], arr[4] only.

psudo code:

NSArray *arr;
for(int i = 0; i < arr.count ; i++)
{
  NSNumber *num1 = [arr objectAtIndex:i];
  for(int j = i + 1; j < arr.count; j++)
     {
       NSNumber *num2 = [arr objectAtIndex:j];
       //compare here num1 with num2
     }
 }