有一个更简单的解决方案，如果你利用NSArray内置的reverseObjectEnumerator方法，以及NSEnumerator的allObjects方法:

NSArray* reversedArray = [[startArray reverseObjectEnumerator] allObjects];

allObjects被记录为返回一个数组，其中的对象还没有被nextObject遍历，顺序如下:

该数组按枚举顺序包含枚举器的所有剩余对象。

一些基准

1. reverseObjectEnumerator allObjects

这是最快的方法:

NSArray *anArray = @[@"aa", @"ab", @"ac", @"ad", @"ae", @"af", @"ag",
        @"ah", @"ai", @"aj", @"ak", @"al", @"am", @"an", @"ao", @"ap", @"aq", @"ar", @"as", @"at",
        @"au", @"av", @"aw", @"ax", @"ay", @"az", @"ba", @"bb", @"bc", @"bd", @"bf", @"bg", @"bh",
        @"bi", @"bj", @"bk", @"bl", @"bm", @"bn", @"bo", @"bp", @"bq", @"br", @"bs", @"bt", @"bu",
        @"bv", @"bw", @"bx", @"by", @"bz", @"ca", @"cb", @"cc", @"cd", @"ce", @"cf", @"cg", @"ch",
        @"ci", @"cj", @"ck", @"cl", @"cm", @"cn", @"co", @"cp", @"cq", @"cr", @"cs", @"ct", @"cu",
        @"cv", @"cw", @"cx", @"cy", @"cz"];

NSDate *methodStart = [NSDate date];

NSArray *reversed = [[anArray reverseObjectEnumerator] allObjects];

NSDate *methodFinish = [NSDate date];
NSTimeInterval executionTime = [methodFinish timeIntervalSinceDate:methodStart];
NSLog(@"executionTime = %f", executionTime);

结果:executionTime = 0.000026

2. 迭代一个reverseObjectEnumerator对象

这要慢1.5到2.5倍:

NSDate *methodStart = [NSDate date];
NSMutableArray *array = [NSMutableArray arrayWithCapacity:[anArray count]];
NSEnumerator *enumerator = [anArray reverseObjectEnumerator];
for (id element in enumerator) {
    [array addObject:element];
}
NSDate *methodFinish = [NSDate date];
NSTimeInterval executionTime = [methodFinish timeIntervalSinceDate:methodStart];
NSLog(@"executionTime = %f", executionTime);

结果:executionTime = 0.000071

3.sortedArrayUsingComparator

这要慢30到40倍(这里没有意外):

NSDate *methodStart = [NSDate date];
NSArray *reversed = [anArray sortedArrayUsingComparator: ^(id obj1, id obj2) {
    return [anArray indexOfObject:obj1] < [anArray indexOfObject:obj2] ? NSOrderedDescending : NSOrderedAscending;
}];

NSDate *methodFinish = [NSDate date];
NSTimeInterval executionTime = [methodFinish timeIntervalSinceDate:methodStart];
NSLog(@"executionTime = %f", executionTime);

结果:executionTime = 0.001100

因此，当谈到速度和易用性时，[[anArray reverseObjectEnumerator] allObjects显然是赢家。

Swift 3语法:

let reversedArray = array.reversed()

我不知道有什么内置方法。 但是，手工编码并不太难。假设你正在处理的数组元素是整数类型的NSNumber对象，'arr'是你想反转的NSMutableArray。

int n = [arr count];
for (int i=0; i<n/2; ++i) {
  id c  = [[arr objectAtIndex:i] retain];
  [arr replaceObjectAtIndex:i withObject:[arr objectAtIndex:n-i-1]];
  [arr replaceObjectAtIndex:n-i-1 withObject:c];
}

因为你从一个NSArray开始，那么你必须先用原始NSArray ('origArray')的内容创建可变数组。

NSMutableArray * arr = [[NSMutableArray alloc] init];
[arr setArray:origArray];

编辑:修正了循环计数中的n -> n/2，并根据布伦特的回答中的建议将NSNumber更改为更通用的id。

或者scala方式:

-(NSArray *)reverse
{
    if ( self.count < 2 )
        return self;
    else
        return [[self.tail reverse] concat:[NSArray arrayWithObject:self.head]];
}

-(id)head
{
    return self.firstObject;
}

-(NSArray *)tail
{
    if ( self.count > 1 )
        return [self subarrayWithRange:NSMakeRange(1, self.count - 1)];
    else
        return @[];
}

Georg Schölly的分类非常好。然而，对于NSMutableArray，当数组为空时，使用NSUIntegers作为索引会导致崩溃。正确的代码是:

@implementation NSMutableArray (Reverse)

- (void)reverse {
    NSInteger i = 0;
    NSInteger j = [self count] - 1;
    while (i < j) {
        [self exchangeObjectAtIndex:i
                  withObjectAtIndex:j];

        i++;
        j--;
    }
}

@end