如何在Objective-C中反转NSArray？

Question

我需要扭转我的态度NSArray.

举个例子:

[1,2,3,4,5] 必须成为: [5,4,3,2,1]

实现这一目标的最佳方法是什么？

Answer 1

如果您利用内置reverseObjectEnumerator方法NSArray和以下allObjects方法,有一个更容易的解决方案NSEnumerator:

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

allObjects记录为返回一个数组,其中包含尚未遍历的对象nextObject,按顺序:

此数组以枚举顺序包含枚举器的所有剩余对象.

Answer 2

为了得到阵列的反转副本,看danielpunkass'溶液使用reverseObjectEnumerator.

要反转可变数组,可以在代码中添加以下类别:

@implementation NSMutableArray (Reverse)

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

        i++;
        j--;
    }
}

@end

Answer 3

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reverseObjectEnumerator

这速度在1.5x到2.5x之间:

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

速度在30x到40x之间(这里没有惊喜):

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

[[anArray reverseObjectEnumerator] allObjects]在速度和轻松方面,明显的赢家也是如此.

Answer 4

DasBoot有正确的方法,但他的代码中有一些错误.这是一个完全通用的代码片段,可以反转任何NSMutableArray:

/* Algorithm: swap the object N elements from the top with the object N 
 * elements from the bottom. Integer division will wrap down, leaving 
 * the middle element untouched if count is odd.
 */
for(int i = 0; i < [array count] / 2; i++) {
    int j = [array count] - i - 1;

    [array exchangeObjectAtIndex:i withObjectAtIndex:j];
}

您可以将其包装在C函数中,或者对于奖励积分,使用类别将其添加到NSMutableArray.(在这种情况下,'array'将变成'self'.)你也可以通过[array count]在循环之前分配变量并使用该变量来优化它,如果你愿意的话.

如果你只有一个常规的NSArray,就无法在适当的位置反转它,因为无法修改NSArrays.但你可以制作一个反向副本:

NSMutableArray * copy = [NSMutableArray arrayWithCapacity:[array count]];

for(int i = 0; i < [array count]; i++) {
    [copy addObject:[array objectAtIndex:[array count] - i - 1]];
}

或者使用这个小技巧在一行中完成:

NSArray * copy = [[array reverseObjectEnumerator] allObjects];

如果您只想向后循环数组,可以使用for/ inloop [array reverseObjectEnumerator],但使用它可能更有效-enumerateObjectsWithOptions:usingBlock::

[array enumerateObjectsWithOptions:NSEnumerationReverse
                        usingBlock:^(id obj, NSUInteger idx, BOOL *stop) {
    // This is your loop body. Use the object in obj here. 
    // If you need the index, it's in idx.
    // (This is the best feature of this method, IMHO.)
    // Instead of using 'continue', use 'return'.
    // Instead of using 'break', set '*stop = YES' and then 'return'.
    // Making the surrounding method/block return is tricky and probably
    // requires a '__block' variable.
    // (This is the worst feature of this method, IMHO.)
}];

(注意:2014年基本更新,基础知识经验还有5年,新的Objective-C功能或两个,以及评论中的一些提示.)

Answer 5

在回顾了上面的其他答案之后,在这里找到了Matt Gallagher的讨论

我建议这个:

NSMutableArray * reverseArray = [NSMutableArray arrayWithCapacity:[myArray count]]; 

for (id element in [myArray reverseObjectEnumerator]) {
    [reverseArray addObject:element];
}

正如马特所说:

在上面的例子中,您可能想知道 - [NSArray reverseObjectEnumerator]是否会在循环的每次迭代中运行 - 可能会降低代码速度.<...>

此后不久,他回答如下:

<...>当for循环开始时,"collection"表达式仅被计算一次.这是最好的情况,因为您可以安全地在"集合"表达式中放置一个昂贵的函数,而不会影响循环的每次迭代性能.

Answer 6

GeorgSchölly的分类非常好.但是,对于NSMutableArray,使用NSUIntegers作为索引会导致数组为空时崩溃.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

Answer 7

反向枚举数组的最有效方法:

使用enumerateObjectsWithOptions:NSEnumerationReverse usingBlock.使用@ JohannesFahrenkrug上面的基准测试,这比[[array reverseObjectEnumerator] allObjects];以下快了8倍:

NSDate *methodStart = [NSDate date];

[anArray enumerateObjectsWithOptions:NSEnumerationReverse usingBlock:^(id obj, NSUInteger idx, BOOL *stop) {
    //
}];

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

Answer 8

NSMutableArray *objMyObject = [NSMutableArray arrayWithArray:[self reverseArray:objArrayToBeReversed]];

// Function reverseArray 
-(NSArray *) reverseArray : (NSArray *) myArray {   
    return [[myArray reverseObjectEnumerator] allObjects];
}