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Suppose I have an array having elements "am","john","rosa","freedom". I want to compare these elements and result would be the word and the size of the longest word. I am using objective C.

user12345
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  • possible duplicate? http://stackoverflow.com/questions/1137839/comparing-arrays-in-objective-c – Adnan Aug 31 '10 at 15:03
  • Could you precise : what kind of array (C style array or NSArray) ? Rem :comparing two element always return a single value (NSOrderDescending, NSOrderAscending or NSOrderedSame). – VdesmedT Aug 31 '10 at 15:18
  • Are `elements` instances of `NSString` or some other class? – falconcreek Aug 31 '10 at 15:34

3 Answers3

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There isn't a "built-in" way of doing this, however you can use NSArray's sortedArrayUsingSelector: and create a category on NSString to provide a lengthCompare: method.

// NSString+LengthCompare.h
#import NSString.h
@interface NSString (LengthComparison)
- (NSComparisonResult)lengthCompare:(NSString *)aString;
@end

// NSString+LengthCompare.m
#import NSString+LengthCompare.h
@implememtation NSString (LengthComparison)
- (NSComparisonResult)lengthCompare:(NSString *)aString
{        
    if ([self length] < [aString length]) {
        return NSOrderedAscending;
    } else if ([self length] > [aString length]) {
        return NSOrderedDescending;
    } else {
        return NSOrderedSame;
    }
}
@end

Now you can sort an of strings in ascending order using lengthCompare:

NSArray *array = [NSArray arrayWithObjects: @"am", @"john", @"rosa", @"freedom", nil];
NSArray *arraySortedByStringLength = [array sortedArrayUsingSelector:@selector(lengthCompare:)];

NString *shortestWord = [[arraySortedByStringLength objectAtIndex:0] retain];
NSLog(@"shortest word, %@ has length %d", shortestWord, [shortestWord length];
[shortestWord release];

NString *longestWord = [[arraySortedByStringLength lastObject] retain];
NSLog(@"Longest word, %@ has length %d", longestWord, [longestWord length];

[longestWord release];
falconcreek
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    If you're only going to do it once, you could also use -[NSArray sortedArrayUsingComparator:], which uses a block instead of a selector. Then you don't have to declare a method that only gets called once. – Alex Martini Aug 31 '10 at 16:29
  • Good alternative however blocks are limited to iOS4 and above. – falconcreek Aug 31 '10 at 16:38
  • tag is [iphone]. on the Mac, blocks were introduced in 10.6 (Snow Leopard) – falconcreek Aug 31 '10 at 17:59
  • updated code to correctly select the longest word given the sortedArrayUsingSelector: returns an array sorted in *ascending* order – falconcreek Sep 01 '10 at 16:16
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Sounds like a classical logic exercise or is it something I miss in your question ?

int longestWordIndex = 0;
NSUInteger longestWordSize = 0;
for (int i=0 ; i<[nameArray count] ; i++) {
  NSString* element = (NSString*)[nameArray objectAtindex:i];
  if([element lenght] > longestWordSize) {
    longestWordSize = [element lenght];
    longestWordIndex = i;
  } 
}
NSLog("Longest word is %@ with size of :%d", [nameArray objectAtIndex:longestWordIndex], longestWordSize);
VdesmedT
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  • well thanx,I know this solution. But this is a lay man solution. If you can tell me any inbuilt methods which is available in objecive C. – user12345 Aug 31 '10 at 15:29
  • Yep, I thought so. You could probably mess up with indexesOfObjectsPassingTest but you will have to call it twice anyway. No other idea – VdesmedT Aug 31 '10 at 15:52
0

I'll add one more approach to the two above -- use a block to do the body of your iteration.

__block NSUInteger longestWordSize = -1; // Make sure at least one object will be longer.
__block NSUInteger longestWordIndex;

[nameArray enumerateObjectsUsingBlock:^(id currentWord, NSUInteger index, BOOL *stop) {
    if ([currentWord length] > longestWordSize) {
        longestWordSize = [currentWord length];
        longestWordIndex = index;
    }
}];

NSLog("Longest word is %@ with size of :%d", [nameArray objectAtIndex:longestWordIndex], longestWordSize);

Edit: The max and index have to be of storage type __block so they can be changed from inside the block.

Alex Martini
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