Question

我想在不同elements中找到arrays的组合。假设我有三个NSArray个对象：

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}];

修改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

使用以下附加功能 a2的每个元素a1的所有元素：

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;
}

你可以通过从一个空数组开始迭代地得到结果将它与你的套装结合起来：

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:@", "]);
}

输出

{  }
{ 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 }

Answer 2

如果您在app环境中有可用的数据库，则可以创建临时表并在它们之间进行交叉连接以获得所需的组合。干杯

不同NSArray对象的组合

2 个答案: