Sort NSArray with array of index numbers

Question

I have an array like this:

NSArray *needSortedArray = @[@"Alex", @"Rachel", @"Mohamad"];

and an array of index like this:

NSArray *indexArray = @[@1, @0, @2];

So the output I want will look like this:

needSortedArray = @[@"Rachel", @"Alex", @"Mohamad"];

How can I do this? Is it possible?

dev gr · Accepted Answer · 2016-04-01 10:51:10Z

5

Try this:

    NSArray *unsortedArray = @[@"Alex", @"Rachel", @"Mohamad"];
    NSArray *indexArray = @[@1, @0, @2];

    NSMutableArray * sortedArray = [[NSMutableArray alloc] initWithCapacity:unsortedArray.count];
    for (NSNumber * num in indexArray)
    {
        [sortedArray addObject:[unsortedArray objectAtIndex:num.integerValue]];
    }

   //now sortedArray has sorted objects.

answered Apr 1, 2016 at 10:51

dev gr

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1 Comment

Andrew Romanov Over a year ago

Good solution if indexes are not equal and in correct indexes range for unsortedArray.

Andrew Romanov · Accepted Answer · 2016-04-01 11:13:09Z

0

Solution that supports any comparable types of indexes:

NSArray<NSString*> *unsortedArray = @[@"Alex", @"Rachel", @"Mohamad", @"Andrew"];
NSArray<NSNumber*> *indexArray = @[@1, @12, @23, @12];

NSParameterAssert([indexArray count] == [unsortedArray count]);
NSArray* sorted = [unsortedArray sortedArrayUsingComparator:^NSComparisonResult(id  _Nonnull obj1, id  _Nonnull obj2) {
    NSNumber* index1 = indexArray[[unsortedArray indexOfObjectIdenticalTo:obj1]];
    NSNumber* index2 = indexArray[[unsortedArray indexOfObjectIdenticalTo:obj2]];

    NSComparisonResult result = [index1 compare:index2];
    return result;
}];

NSLog(@"Sorted array: %@", sorted);

answered Apr 1, 2016 at 11:13

Andrew Romanov

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