不同NSArray对象的组合

Question

我想找到elements不同的组合arrays.假设我有三个NSArray对象:

NSArray *set1 = [NSArray arrayWithObjects:@"A",@"B",@"C", nil];
NSArray *set2 = [NSArray arrayWithObjects:@"a",@"b", nil];
NSArray *set3 = [NSArray arrayWithObjects:@"1",nil];

现在所需的答案是在数组之后

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}];

编辑 目前我已经完成了以下代码,我可以得到两个长度的组合.

    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);
                }

            }

        }

    }

任何帮助???

Answer 1

使用以下函数,该函数将所有元素追加a2到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;
}

您可以通过从空数组开始并将其与您的集合相结合来迭代地获得结果:

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);

显示结果:

for (NSArray *item in result) {
    NSLog(@"{ %@ }", [item componentsJoinedByString:@", "]);
}

产量

{  }
{ 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 }