# 不同nsArray对象的组合[英] Combinations of different NSArray objects

### 问题描述

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

```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}];
```

```    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++) {
}

NSMutableArray *singleCombinations = combinations[0];

for (NSArray *set in allSets) {
}

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);
}

}

}

}
```

## 推荐答案

```NSArray *combinations(NSArray *a1, NSArray *a2)
{
NSMutableArray *result = [NSMutableArray array];
for (NSArray *elem1 in a1) {
for (id elem2 in a2) {
}
}
return result;
}
```

```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);
```

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

```{  }
{ 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 }
```