生成隨機字母串的有效方法？

Question

我想要隨機化字母表中所有字符的字符串。現在，我創建了26個字符的可變數組，使用exchangeObjectAtIndex：方法對它們進行混洗，然後將每個字符添加到我返回的字符串中。

必須有更好的方法來做到這一點。這裏是我的代碼：

- (NSString *)shuffledAlphabet { 
    NSMutableArray * shuffledAlphabet = [NSMutableArray arrayWithArray:@[@"A",@"B",@"C",@"D",@"E",@"F",@"G",@"H",@"I",@"J",@"K",@"L",@"M",@"N",@"O",@"P",@"Q",@"R",@"S",@"T",@"U",@"V",@"W",@"X",@"Y",@"Z"]]; 

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

    NSString *string = [[NSString alloc] init]; 
    for (NSString *letter in shuffledAlphabet) { 
     string = [NSString stringWithFormat:@"%@%@",string,letter]; 
    } 

    return string; 
} 

Answer 1

下面是一個有效的Fisher-Yates shuffle，適應你的使用情況：

- (NSString *)shuffledAlphabet { 
    NSString *alphabet = @"ABCDEFGHIJKLMNOPQRSTUVWXYZ"; 

    // Get the characters into a C array for efficient shuffling 
    NSUInteger numberOfCharacters = [alphabet length]; 
    unichar *characters = calloc(numberOfCharacters, sizeof(unichar)); 
    [alphabet getCharacters:characters range:NSMakeRange(0, numberOfCharacters)]; 

    // Perform a Fisher-Yates shuffle 
    for (NSUInteger i = 0; i < numberOfCharacters; ++i) { 
     NSUInteger j = (arc4random_uniform(numberOfCharacters - i) + i); 
     unichar c = characters[i]; 
     characters[i] = characters[j]; 
     characters[j] = c; 
    } 

    // Turn the result back into a string 
    NSString *result = [NSString stringWithCharacters:characters length:numberOfCharacters]; 
    free(characters); 
    return result; 
} 

Answer 2

Generating random numbers in Objective-C對此是否有幫助？ *除以26產生的隨機數 並採取提醒 *索引數組[提醒]

Answer 3

你可以挑選從（其餘的）字母隨機元素，當你建立你的字符串，而不是第一次洗牌它的：

NSMutableArray *alphabet = [NSMutableArray arrayWithObjects:@"A",@"B",@"C",@"D",@"E",@"F",@"G",@"H",@"I",@"J",@"K",@"L",@"M",@"N",@"O",@"P",@"Q",@"R",@"S",@"T",@"U",@"V",@"W",@"X",@"Y",@"Z", nil]; 
NSMutableString *result = [NSMutableString string]; 
NSUInteger numberOfLetters = alphabet.count; 
for (NSUInteger i = 0; i < numberOfLetters; i++) { 
    int n = arc4random() % alphabet.count; 
    [result appendString:[alphabet objectAtIndex:n]]; 
    [alphabet removeObjectAtIndex:n]; 
} 
NSLog(@"%@", result); 

這使代碼更短一些。還請注意，使用NSMutableString比每次添加一封信時創建新的NSString更有效。

Answer 4

這是執行正常洗牌字母代更有效的方式。

- (NSString *)shuffledAlphabet 
{ 
    const NSUInteger length = 'Z' - 'A' + 1; 
    unichar alphabet[length]; 
    alphabet[0] = 'A'; 

    for (NSUInteger i = 1; i < length; i++) 
    { 
     NSUInteger j = arc4random_uniform((uint32_t)i + 1); 
     alphabet[i] = alphabet[j]; 
     alphabet[j] = 'A' + i; 
    } 
    return [NSString stringWithCharacters:alphabet length:length]; 
} 

它使用費耶茨洗牌的「由內而外」的版本，並通過與arc4random_uniform產生僞隨機數可以避免MODULA偏差。此外，它需要一個單獨的分配，因爲所有的排列都在一個臨時緩衝區中執行。