Question

我正在尝试用6个不同字符中的4个填充数组，其中数组中的每个字符都不同。放入数组的字符基于switch语句。然后我想回到数组，以确保我刚刚放置的角色不在数组中，如果是，我必须放入一个不同的角色。我该怎么做呢？我不想重新填充整个阵列，只是这一点。

这是我到目前为止的代码，评论包括：

-(void) fillarray{

for (int i=0; i<4; i++)
{
    int randomNum = arc4random() % 6; // generates a random number

          switch (randomNum) // based on the number generated, it choses what color will be placed into the slot
    {
        case 1:
            colorarray[i] ='r';
            break;

        case 2:
            colorarray[i] ='o';
            break;

        case 3:
            colorarray[i] ='y';
            break;

        case 4:
            colorarray[i] ='g';
            break;
        case 5:
            colorarray[i] = 'b';

        case 6:
            colorarray[i] = 'p';

        default:
            break;
    }

    for (int j=0; j<4; j++) // runs through the array to ensure that the recently placed color does not match previous placed colors
    {
        while (j /= i)

            if (colorarray[i]==colorarray[j]) // if the color is already in the array
                                            //try again to place another color in this location

    }

Answer 1

首先你拿一个数组（你也可以使用NSArray，我想你想要使用C风格的数组）来提取所有可能的字符：

char characters[]={'r','o','y','g','b','p'};
int length=6;

然后每次提取一个字符时，减少长度变量并交换最后一个提取的字符，以确保它不会再次使用：

int randomNum = arc4random() % length;
< put characters[randomNum] somewhere>
char temp=characters[randomNum];
characters[randomNum]=characters[length-1];
characters[length-1]=temp;
length--;

PS：％操作数返回商的余数，因此数字％N永远不会是N，它的范围从0到N-1。

Answer 2

我会根据Fisher-Yates shuffling algorithm

使用此功能

NSArray * array = @[ @"r", @"o", @"y", @"g", @"b", @"p" ] ;
array = [ [ array shuffledArray ] subarrayWithRange:(NSRange){ .length = 4 } ] ;

通过以下方式将-shuffledArray添加到NSArray中：

@implementation NSArray ( Shuffling )

-(NSArray*)shuffledArray
{
    NSMutableArray * result = [ NSMutableArray arrayWithCapacity:self.count ] ;
    [ result addObject:self[0] ] ;
    for( NSUInteger index=1, count = self.count; index < count; ++index )
    {
        int j = arc4random_uniform( (unsigned int)index ) ;
        [ result addObject:result[j] ] ;
        [ result replaceObjectAtIndex:j withObject:self[ index ] ] ;
    }

    return result ;
}

@end

Answer 3

该函数应如下所示：

    int found = 1;
    for (int i=0; i<4; i++)
    {
        while(found) {
           int randomNum = arc4random() % 6;
           switch (randomNum) 
           {
              case 1:
                colorarray[i] ='r';
                break;

              case 2:
                colorarray[i] ='o';
                break;

             case 3:
               colorarray[i] ='y';
               break;

             case 4:
               colorarray[i] ='g';
               break;
             case 5:
               colorarray[i] = 'b';
               break;

             case 6:
               colorarray[i] = 'p';

             default:
               break;
           }
           found = 0;
           for (int j = i-1; i >= 0; j--) {
              if (colorarray[i] == colorarray[j])
                 found = 1;
           }
        }
 }

Answer 4

此外，如果您不想使用switch或（基本上）通过递减长度变量来删除从数组中收到的字符，您可以始终执行此操作：

char colorarray[5] = {0,0,0,0,'\0'};
char possibleCharacters[6] = {'r', 'o', 'y', 'g', 'b', 'p'};
BOOL isNotUnique = NO;

for(int i = 0; i < 4; i++){
    do{
        colorarray[i] = possibleCharacters[arc4random() % 6];
        for(int j = i - 1; j > -1; --j){
            if(colorarray[i] == colorarray[j]){
                isNotUnique = YES;
                break;
            }
            else isNotUnique = NO;
        }
    }while(isNotUnique);
}

NSLog(@"%s", colorarray);

获取字符的实际循环基本上是：

1    for each spot in the colors arrays
2        generate a random number and use it to get a character
3        check to see if the character is identical to any spot before this one
4        if our new character matched any previous one, repeat from 2 on

Answer 5

这是另一种改组数组的方法（获得随机的前X个元素）。基本上你做Ramy Alzury和nielsbot的回应。算法是这样的。

Do this 10 times the length of the array
    swap the last element with a random element chosen from the array

最后这应该产生一个随机洗牌的数组（你将它洗牌10次）

这是目标C中的示例函数。

-(NSArray *) shuffleArray:(NSArray *) arr{
    NSMutableArray * mArr = [NSMutableArray arrayWithArray:arr];
    for(int i=0;i<arr.count*10;i++){
        int num = arc4random()%arr.count;
        NSString * temp = (NSString *)[mArr objectAtIndex:arr.count-1];
        [mArr replaceObjectAtIndex: arr.count-1 withObject:[mArr objectAtIndex:num]]; 
        [mArr replaceObjectAtIndex:num withObject:temp];
    }
    return [NSArray arrayWithArray:mArr];
}

使用从列表中选择的随机字符填充数组，而不重复

5 个答案: