Получите NSIndexSet от NSArray

It might be useful to implement it using a set to specify the objects to find, such as:

- (NSIndexSet *) indicesOfObjectsInSet: (NSSet *) set
{
    if ( [set count] == 0 )
        return ( [NSIndexSet indexSet] );

    NSMutableIndexSet * indices = [NSMutableIndexSet indexSet];

    NSUInteger index = 0;
    for ( id obj in self )
    {
        if ( [set containsObject: obj] )
            [indices addIndex: index];

        index++;
    }

    return ( [[indices copy] autorelease] );
}

This requires visiting every object in the array, but at least only does so once and makes use of fast enumeration while doing so. Using an NSSet and testing each object in the array against that set is also much faster than testing for inclusion in an array.

There's a potential optimization here, but it would break in the case where a single object is stored in the receiving array multiple times:

if ( [set containsObject: obj] )
{
    [indices addIndex: index];
    if ( [indices count] == [set count] )
        break;
}

That way if you're scanning a 20'000-item array for two objects and they're both inside the first ten, you'll be able to avoid scanning the other 19'990 objects in the array. As I said though, that doesn't help if the array contains duplicates, because it'll stop as soon as it's found 2 indices (even if they both point to the same object).

Having said that, I agree with Mike's comment above. Chances are you're setting yourself up for some pain come optimization-time. It may be worth thinking about different data types; for instance, while NSArray seems the most logical choice for a simple flat container, if you don't actually need the ordering information it's better to use an NSSet instead; this has the added advantage that it won't store the same object (calculated using -isEqual:) twice. If you do want to keep track of duplicates, but don't need ordering, you can use NSCountedSet, which behaves as NSSet except it keeps track of how many times each objects has been added/removed without actually storing duplicates.