如何在不硬核索引的情况下擦除父数组中子数组的重复项

Question

我有两个数组，

NSArray *First=@[@"a", @"b", @"c",@"d", @"e", @"f",@"g", @"h", @"i", @"j", @"k"];
NSArray *Second=@[@"a",@"d", @"e",@"g",@"i"];

现在在

 NSMutableArray *Result=[[NSMutableArray alloc]init];

当我使用NSLog结果数组时，我希望输出为

(bcfhjk)

表示结果数组从第一个数组中删除第二个数组中的元素，且条件为no out using removeObjectAtIndex Method

谢谢提前。

Answer 1

for (id object in First)
    if (![Second containsObject:object]) [Result addObject:object];

Answer 2

下面是在不使用removeObjectAtIndex方法的情况下获得所需输出的代码。试试这个:

NSArray *First=@[@"a", @"b", @"c",@"d", @"e", @"f",@"g", @"h", @"i", @"j", @"k"];
    NSArray *Second=@[@"a",@"d", @"e",@"g",@"i"];
    NSMutableArray *result=[[NSMutableArray alloc]init];

    for (NSString *tempChar in First) {
        if (![Second containsObject:tempChar]) {
            [result addObject:tempChar];
            NSLog(@"result arr :%@",result);
        }
    }

Answer 3

// *** Most Efficient way to achieve with just 3 Lines of code ***
// *** `NSMutableSet` will do the job for you. ***
// *** method `minusSet` performs subtraction operation between two given sets. ***
NSMutableSet *first = [[NSMutableSet alloc] initWithArray:@[@"a", @"b", @"c",@"d", @"e", @"f",@"g", @"h", @"i", @"j", @"k"]];
NSMutableSet *second = [[NSMutableSet alloc] initWithArray:@[@"a",@"d", @"e",@"g",@"i"]];
[first minusSet:second];
NSLog(@"%@",first);

Answer 4

如果没有循环，可以使用NSSet存档

NSMutableSet *firstSet1 = [NSMutableSet setWithArray: First];
NSSet *secondSet2 = [NSSet setWithArray: Second];
[firstSet1 minusSet: secondSet2];
NSArray * Result = [firstSet1 allObjects];