Question

目标是生成长度为NSString的字符，并将每个字符串分配给一个数组。我陷入了我需要用算法做的事情以获得正确的结果。这是样本。我得到的结果是相同的随机生成的字符串添加到我的数组26次而不是添加26个不同的字符串。

我考虑过声明26个不同的NSStrings并将算法的每个结果分配给每个字符串，但这似乎效率低下。谢谢你的帮助。

NSMutableString *string = @"expert";
NSUInteger strLength = [string length];
NSString *letterToAdd;
NSString *finishedWord;
NSMutableString *randomString = [NSMutableString stringWithCapacity: strLength];
NSMutableArray *randomArray = [[NSMutableArray alloc] init];

NSArray *charArray = [[NSArray alloc] initWithObjects: @"a", @"b", @"c", @"d", 
                         @"e", @"f", @"g", @"h", @"i", @"j", @"k", @"l", @"m", 
                         @"o", @"p", @"q", @"r", @"s", @"t", @"u", @"v", @"w", 
                         @"x", @"y", @"z", nil];

for (int a = 0; a < 26; a++) {
  for (int i = 0; i < strLength; i++) {

    letterToAdd = [charArray objectAtIndex: arc4random() % [charArray count]];
    if([randomString length] < strLength) {
      [randomString insertString: letterToAdd atIndex: i];
    }

    finishedWord = randomString;
  }

  [randomArray addObject: finishedWord];   
}

NSLog(@"Random Array count %i, contents: %@", [randomArray count], randomArray);

Answer 1

我将如何做到这一点：

#import "NSString+Shuffle.h"
NSString * string = @"expert";
NSUInteger strLength = [string length];
NSString * alphabet = @"abcdefghijklmnopqrstuvwxyz";
NSMutableSet * randomWords = [NSMutableSet set];

while ([randomWords count] < 26) {
  NSString * newWord = [alphabet shuffledString];
  newWord = [newWord substringToIndex:strLength];
  [randomArray addObject:newWord];
}
NSLog(@"Random set count %d, contents: %@", [randomWords count], randomWords);

然后，您需要NSString上定义shuffledString的类别。此方法将简单地获取字符串中的字符并随机重新排列它们。 Google可以很容易地找到体面的随机播放算法。

我希望你能够了解它的工作原理。我做的唯一修改是使用NSSet而不是NSArray，以及循环的条件是什么。消除了重复随机单词的（微小）可能性。

编辑：因为我感觉很慷慨，所以这是一个基本的shuffledString实施：

//NSString+Shuffle.h
@interface NSString (ShuffleAdditions)

- (NSString *) shuffledString;

@end

//NSString+Shuffle.m
#import "NSString+Shuffle.h"

@implementation NSString (ShuffleAdditions)

- (NSString *) shuffledString {
  NSMutableString * shuffled = [self mutableCopy];
  NSUInteger length = [shuffled length];
  for (int i = 0; i < (4*length); ++i) {
    NSString * randomChar = [shuffled subStringWithRange:NSMakeRange(arc4random() % (length-1), 1)];
    [shuffled appendString:randomChar];
  }
  return [shuffled autorelease];
}
@end

Answer 2

每次都应该创建一个新的randomString：

NSMutableString *string = @"expert";
NSUInteger strLength = [string length];
NSString *letterToAdd;
NSString *finishedWord;
//NSMutableString *randomString = [NSMutableString stringWithCapacity: strLength];
NSMutableArray *randomArray = [[NSMutableArray alloc] init];

NSArray *charArray = [[NSArray alloc] initWithObjects: @"a", @"b", @"c", @"d", @"e", @"f",
                      @"g", @"h", @"i", @"j", @"k", @"l", @"m", @"o", @"p", @"q", @"r", @"s",
                      @"t", @"u", @"v", @"w", @"x", @"y", @"z", nil];

for (int a = 0; a < 26; a++) {
    NSMutableString *randomString = [NSMutableString stringWithCapacity: strLength];

    for (int i = 0; i < strLength; i++) {

            letterToAdd = [charArray objectAtIndex: arc4random() % [charArray count]];
                //if([randomString length] < strLength) {
                    [randomString insertString: letterToAdd atIndex: i];
                //}

        //finishedWord = randomString;
    }

    //[randomArray addObject: finishedWord];
    [randomArray addObject: randomString];
}

NSLog(@"Random Array count %i, contents: %@", [randomArray count], randomArray);

Answer 3

每次循环时都会向阵列添加相同的对象，并在你去的时候覆盖它。你提到过：

我考虑过声明26个不同的NSStrings并分配每个NSStrings 从算法到每个字符串的结果......

这确实是你需要做的。将randomString的初始化移动到循环中将解决您的问题（在每次循环迭代中获得新的NSMutableString，而不是使用单个对象）。将randomString的定义更改为简单类型定义：

NSMutableString *randomString;

然后在你的外循环中添加以下行：

randomString = [NSMutableString stringWithCapacity:strLength];

您不需要更改任何其余代码。

算法将随机生成的NSStrings添加到NSMutableArray

3 个答案: