ios - NSArray 排序纵横交错

Question

我有一个数组

@["A","B","C","D","E","F","G"]

我想像下面这样交叉排序

@["A","G","B","F","C","E","D"]

使最后一个项目每 2 次滑到前面。

Answer 1

我看你什么都没试过。如果是因为您还没有找到任何算法，或者您想不出一个算法，那么这段代码可能会有所帮助：

@autoreleasepool
{
    BOOL tail= NO; // To know if you should remove from array's tail or head.
    NSMutableArray* array=[NSMutableArray arrayWithArray: @[@"A",@"B",@"C",@"D",@"E",@"F",@"G"] ]; // The unsorted array.
    // This will contain sorted objects:  
    NSMutableArray* sorted=[[NSMutableArray alloc]initWithCapacity: array.count];
    // The algorithm will end when array will be empty:  
    while(array.count)
    {
        NSUInteger index= tail? array.count-1:0; // I decide the index of the object
                                                 // to remove.
        // The removed object will be added to the sorted array, so that it will
        // contain the object on head, then on tail, then again on head, and so on...
        id object= array[index];
        [sorted addObject: object];
        [array removeObjectAtIndex: index];
        tail= !tail;
    }
    NSLog(@"%@",sorted);
}

Answer 2

这可以这样做：

将数组分成两半。

对它们进行排序。

再次合并它们以形成您的结果。在这里您需要迭代替代方案，即 step+=2

编辑：运行代码如下

NSArray *array=@[@"A",@"B",@"C",@"D",@"E",@"F",@"G"];
NSArray *halfLeft=[array subarrayWithRange:NSMakeRange(0, array.count/2+1)];
NSMutableArray *halfRight=[NSMutableArray arrayWithArray:array];
[halfRight removeObjectsInArray:halfLeft];
NSMutableArray *finalAray=[[NSMutableArray alloc]initWithArray:halfLeft];
for (NSInteger i=0, index=1; i<halfRight.count; i++, index+=2) {
    [finalAray insertObject:halfRight[halfRight.count-1-i] atIndex:index];
}
NSLog(@"%@",finalAray);

Answer 3

您可以使用一个简单的循环来完成，将对象添加到输出数组中；

NSArray *input = @[@"A",@"B",@"C",@"D",@"E",@"F",@"G",@"H"];
NSMutableArray *output = [[NSMutableArray alloc] init];

// Quickly add all except possibly the middle one, makes the loop simple
for(int i=0; i<input.count/2; i++)
{
    [output addObject:input[i]];
    [output addObject:input[input.count-i-1]];
}

// If there is an odd number of items, just add the last one separately
if(input.count%2) 
    [output addObject:input[input.count/2]];