问题
I want to find the combinations of the elements
in diffrent arrays
. Let say I've three NSArray
objects as:
NSArray *set1 = [NSArray arrayWithObjects:@"A",@"B",@"C", nil];
NSArray *set2 = [NSArray arrayWithObjects:@"a",@"b", nil];
NSArray *set3 = [NSArray arrayWithObjects:@"1",nil];
Now the required answers is following arrays
NSArray *combinations = [{A},{B},{C},{a},{b},{1},{A,a},{A,b},{A,1},{B,a},{B,b},{B,1},{a,1},{b,1},{A,a,1},{A,b,1},{B,a,1},{B,b,1},{C,a,1},{C,b,1}];
Edit Currently I've did the following code and I'm able to get combinations of two length.
NSArray *set1 = [NSArray arrayWithObjects:@"A",@"B",@"C", nil];
NSArray *set2 = [NSArray arrayWithObjects:@"a",@"b", nil];
NSArray *set3 = [NSArray arrayWithObjects:@"1",nil];
NSArray *allSets = [NSArray arrayWithObjects:set1,set2,set3,nil];
NSMutableArray *combinations = [NSMutableArray new];
for (int index = 0; index < allSets.count; index++) {
[combinations addObject:[NSMutableArray array]];
}
NSMutableArray *singleCombinations = combinations[0];
for (NSArray *set in allSets) {
[singleCombinations addObjectsFromArray:set];
}
for (int outerIndex = 0; outerIndex < allSets.count-1; outerIndex++) {
NSArray *set = allSets[outerIndex];
for (id object1 in set) {
for (int innerIndex = outerIndex+1; innerIndex<allSets.count; innerIndex++) {
NSArray *nextSet = allSets[innerIndex];
for (id object2 in nextSet) {
NSString *combi = [NSString stringWithFormat:@"%@%@",object1,object2];
NSLog(@"%@",combi);
}
}
}
}
Any help???
回答1:
Using the following function, which appends
all elements of a2
to each element of a1
:
NSArray *combinations(NSArray *a1, NSArray *a2)
{
NSMutableArray *result = [NSMutableArray array];
for (NSArray *elem1 in a1) {
[result addObject:elem1];
for (id elem2 in a2) {
[result addObject:[elem1 arrayByAddingObject:elem2]];
}
}
return result;
}
you can get the result iteratively by starting with an empty array and combining that with your sets:
NSArray *set1 = @[@"A", @"B", @"C"];
NSArray *set2 = @[@"a", @"b"];
NSArray *set3 = @[@"1"];
NSArray *result = @[@[]];
result = combinations(result, set1);
result = combinations(result, set2);
result = combinations(result, set3);
Show the result:
for (NSArray *item in result) {
NSLog(@"{ %@ }", [item componentsJoinedByString:@", "]);
}
Output
{ } { 1 } { a } { a, 1 } { b } { b, 1 } { A } { A, 1 } { A, a } { A, a, 1 } { A, b } { A, b, 1 } { B } { B, 1 } { B, a } { B, a, 1 } { B, b } { B, b, 1 } { C } { C, 1 } { C, a } { C, a, 1 } { C, b } { C, b, 1 }
回答2:
If you have a database available in you app enviromnment you could create temp tables and make cross joins between them in order to get the required combinations. Cheers
来源:https://stackoverflow.com/questions/20048824/combinations-of-different-nsarray-objects