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
.
views:
87answers:
3
A:
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
2010-08-31 15:25:08
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.
alis
2010-08-31 15:29:56
Yep, I thought so. You could probably mess up with indexesOfObjectsPassingTest but you will have to call it twice anyway. No other idea
VdesmedT
2010-08-31 15:52:49
+1
A:
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
2010-08-31 16:23:44
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
2010-08-31 16:29:43
tag is [iphone]. on the Mac, blocks were introduced in 10.6 (Snow Leopard)
falconcreek
2010-08-31 17:59:24
updated code to correctly select the longest word given the sortedArrayUsingSelector: returns an array sorted in *ascending* order
falconcreek
2010-09-01 16:16:54
A:
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
2010-08-31 16:43:52