改组NSMutableArray的最佳方法是什么？

如果您有个NSMutableArray，您如何随机洗牌？

（我对此有自己的答案，发布在下面，但是我是可可的新手，我想知道是否有更好的方法。）

更新：正如@Mukesh所指出的那样，从iOS 10+和m​​acOS 10.12+开始，有一种-[NSMutableArray shuffledArray]方法可以用于随机播放。有关详细信息，请参见https://developer.apple.com/documentation/foundation/nsarray/1640855-shuffledarray?language=objc。（但是请注意，这将创建一个新数组，而不是将元素改组到位。）

您不需要swapObjectAtIndex方法。exchangeObjectAtIndex：withObjectAtIndex：已经存在。

我通过在NSMutableArray中添加一个类别来解决此问题。

编辑：删除不必要的方法，感谢拉德的回答。

编辑：更改(arc4random() % nElements)为arc4random_uniform(nElements)感谢格雷戈里·戈尔佐夫（Gregory Goltsov）的回答以及miho和blahdiblah的评论

编辑：循环改进，感谢罗恩的评论

编辑：添加了检查数组不为空，这要感谢Mahesh Agrawal的评论

//  NSMutableArray_Shuffling.h

#if TARGET_OS_IPHONE
#import <UIKit/UIKit.h>
#else
#include <Cocoa/Cocoa.h>
#endif

// This category enhances NSMutableArray by providing
// methods to randomly shuffle the elements.
@interface NSMutableArray (Shuffling)
- (void)shuffle;
@end


//  NSMutableArray_Shuffling.m

#import "NSMutableArray_Shuffling.h"

@implementation NSMutableArray (Shuffling)

- (void)shuffle
{
    NSUInteger count = [self count];
    if (count <= 1) return;
    for (NSUInteger i = 0; i < count - 1; ++i) {
        NSInteger remainingCount = count - i;
        NSInteger exchangeIndex = i + arc4random_uniform((u_int32_t )remainingCount);
        [self exchangeObjectAtIndex:i withObjectAtIndex:exchangeIndex];
    }
}

@end

由于我无法发表评论，因此我认为我会做出完整的回应。我以多种方式修改了克里斯托弗·约翰逊（Kristopher Johnson）对我的项目的实现（确实试图使其尽可能简洁），其中之一是arc4random_uniform()因为它避免了模偏差。

// NSMutableArray+Shuffling.h
#import <Foundation/Foundation.h>

/** This category enhances NSMutableArray by providing methods to randomly
 * shuffle the elements using the Fisher-Yates algorithm.
 */
@interface NSMutableArray (Shuffling)
- (void)shuffle;
@end

// NSMutableArray+Shuffling.m
#import "NSMutableArray+Shuffling.h"

@implementation NSMutableArray (Shuffling)

- (void)shuffle
{
    NSUInteger count = [self count];
    for (uint i = 0; i < count - 1; ++i)
    {
        // Select a random element between i and end of array to swap with.
        int nElements = count - i;
        int n = arc4random_uniform(nElements) + i;
        [self exchangeObjectAtIndex:i withObjectAtIndex:n];
    }
}

@end

如果导入GameplayKit，则有一个shuffledAPI：

https://developer.apple.com/reference/foundation/nsarray/1640855-shuffled

let shuffledArray = array.shuffled()

略有改进和简洁的解决方案（与最佳答案相比）。

该算法是相同的，在文献中被描述为“ Fisher-Yates shuffle ”。

在Objective-C中：

@implementation NSMutableArray (Shuffle)
// Fisher-Yates shuffle
- (void)shuffle
{
    for (NSUInteger i = self.count; i > 1; i--)
        [self exchangeObjectAtIndex:i - 1 withObjectAtIndex:arc4random_uniform((u_int32_t)i)];
}
@end

在Swift 3.2和4.x中：

extension Array {
    /// Fisher-Yates shuffle
    mutating func shuffle() {
        for i in stride(from: count - 1, to: 0, by: -1) {
            swapAt(i, Int(arc4random_uniform(UInt32(i + 1))))
        }
    }
}

在Swift 3.0和3.1中：

extension Array {
    /// Fisher-Yates shuffle
    mutating func shuffle() {
        for i in stride(from: count - 1, to: 0, by: -1) {
            let j = Int(arc4random_uniform(UInt32(i + 1)))
            (self[i], self[j]) = (self[j], self[i])
        }
    }
}

注意：从iOS10使用，可以在Swift中提供更简洁的解决方案GameplayKit。

注意：还提供了一种用于不稳定混洗的算法（如果count> 1，则所有位置都必须更改）

这是改组NSArrays或NSMutableArrays的最简单，最快的方法（对象拼图是NSMutableArray，它包含拼图对象。我已经添加到拼图对象变量索引中，该索引指示数组的初始位置）

int randomSort(id obj1, id obj2, void *context ) {
        // returns random number -1 0 1
    return (random()%3 - 1);    
}

- (void)shuffle {
        // call custom sort function
    [puzzles sortUsingFunction:randomSort context:nil];

    // show in log how is our array sorted
        int i = 0;
    for (Puzzle * puzzle in puzzles) {
        NSLog(@" #%d has index %d", i, puzzle.index);
        i++;
    }
}

日志输出：

 #0 has index #6
 #1 has index #3
 #2 has index #9
 #3 has index #15
 #4 has index #8
 #5 has index #0
 #6 has index #1
 #7 has index #4
 #8 has index #7
 #9 has index #12
 #10 has index #14
 #11 has index #16
 #12 has index #17
 #13 has index #10
 #14 has index #11
 #15 has index #13
 #16 has index #5
 #17 has index #2

您也可以将obj1与obj2进行比较，并确定要返回的可能值是：

• NSOrderedAscending = -1
• NSOrderedSame = 0
• NSOrderedDescending = 1

有一个很好的受欢迎的库，该库包含此方法，在GitHub中称为SSToolKit。文件NSMutableArray + SSToolkitAdditions.h包含随机播放方法。您也可以使用它。其中，似乎有很多有用的东西。

该库的主页在这里。

如果使用此代码，则代码将如下所示：

#import <SSCategories.h>
NSMutableArray *tableData = [NSMutableArray arrayWithArray:[temp shuffledArray]];

该库还具有一个Pod（请参阅CocoaPods）

在iOS 10中，您可以使用GameplayKit中的NSArrayshuffled()。这是Swift 3中Array的助手：

import GameplayKit

extension Array {
    @available(iOS 10.0, macOS 10.12, tvOS 10.0, *)
    func shuffled() -> [Element] {
        return (self as NSArray).shuffled() as! [Element]
    }
    @available(iOS 10.0, macOS 10.12, tvOS 10.0, *)
    mutating func shuffle() {
        replaceSubrange(0..<count, with: shuffled())
    }
}

如果元素有重复。

例如数组：AAABB或BBAAA

唯一的解决方案是：ABABA

sequenceSelected 是一个NSMutableArray，它存储obj类的元素，这些元素是指向某些序列的指针。

- (void)shuffleSequenceSelected {
    [sequenceSelected shuffle];
    [self shuffleSequenceSelectedLoop];
}

- (void)shuffleSequenceSelectedLoop {
    NSUInteger count = sequenceSelected.count;
    for (NSUInteger i = 1; i < count-1; i++) {
        // Select a random element between i and end of array to swap with.
        NSInteger nElements = count - i;
        NSInteger n;
        if (i < count-2) { // i is between second  and second last element
            obj *A = [sequenceSelected objectAtIndex:i-1];
            obj *B = [sequenceSelected objectAtIndex:i];
            if (A == B) { // shuffle if current & previous same
                do {
                    n = arc4random_uniform(nElements) + i;
                    B = [sequenceSelected objectAtIndex:n];
                } while (A == B);
                [sequenceSelected exchangeObjectAtIndex:i withObjectAtIndex:n];
            }
        } else if (i == count-2) { // second last value to be shuffled with last value
            obj *A = [sequenceSelected objectAtIndex:i-1];// previous value
            obj *B = [sequenceSelected objectAtIndex:i]; // second last value
            obj *C = [sequenceSelected lastObject]; // last value
            if (A == B && B == C) {
                //reshufle
                sequenceSelected = [[[sequenceSelected reverseObjectEnumerator] allObjects] mutableCopy];
                [self shuffleSequenceSelectedLoop];
                return;
            }
            if (A == B) {
                if (B != C) {
                    [sequenceSelected exchangeObjectAtIndex:i withObjectAtIndex:count-1];
                } else {
                    // reshuffle
                    sequenceSelected = [[[sequenceSelected reverseObjectEnumerator] allObjects] mutableCopy];
                    [self shuffleSequenceSelectedLoop];
                    return;
                }
            }
        }
    }
}

NSUInteger randomIndex = arc4random() % [theArray count];

克里斯托弗·约翰逊（Kristopher Johnson）的回答很不错，但这并不是完全随机的。

给定一个由2个元素组成的数组，此函数将始终返回反向数组，因为您将在其余索引上生成随机范围。更准确的shuffle()功能是

- (void)shuffle
{
   NSUInteger count = [self count];
   for (NSUInteger i = 0; i < count; ++i) {
       NSInteger exchangeIndex = arc4random_uniform(count);
       if (i != exchangeIndex) {
            [self exchangeObjectAtIndex:i withObjectAtIndex:exchangeIndex];
       }
   }
}

编辑：这是不正确的。仅供参考，我没有删除此帖子。请参阅有关此方法不正确的原因的评论。

简单的代码在这里：

- (NSArray *)shuffledArray:(NSArray *)array
{
    return [array sortedArrayUsingComparator:^NSComparisonResult(id obj1, id obj2) {
        if (arc4random() % 2) {
            return NSOrderedAscending;
        } else {
            return NSOrderedDescending;
        }
    }];
}