在Swift中遇到隨機數問題

Question

我正在嘗試生成隨機的背景色，以在用戶按下屏幕上的按鈕時顯示給用戶。 我有一個不可變的UIColor項數組，並且已經制作了一個可變副本來進行操作。 當生成隨機顏色時，然后將該顏色返回並從數組的可變副本中刪除，以防止連續顯示相同顏色，直到顯示所有顏色為止。 這應該一直發生到數組的計數為0，然后重新創建數組以重復該過程。 但是，當我深入到2到0之間的數組時，循環似乎變成了無限循環。 我的代碼（操場文件）中缺少什么邏輯？

var currentColorIndexNumber = 0
var newColorIndexNumber = 0

let colorsArray = [
    UIColor(red: 90/255.0, green: 187/255.0, blue: 181/255.0, alpha: 1.0), //teal color
    UIColor(red: 222/255.0, green: 171/255.0, blue: 66/255.0, alpha: 1.0), //yellow color
    UIColor(red: 223/255.0, green: 86/255.0, blue: 94/255.0, alpha: 1.0), //red color
    UIColor(red: 239/255.0, green: 130/255.0, blue: 100/255.0, alpha: 1.0), //orange color
    UIColor(red: 77/255.0, green: 75/255.0, blue: 82/255.0, alpha: 1.0), //dark color
    UIColor(red: 105/255.0, green: 94/255.0, blue: 133/255.0, alpha: 1.0), //purple color
    UIColor(red: 85/255.0, green: 176/255.0, blue: 112/255.0, alpha: 1.0), //green color
]

var mutableColorsArray: [AnyObject] = colorsArray

func randomNumber() -> Int {
    // avoid repeating random integers
    while currentColorIndexNumber == newColorIndexNumber {
        var unsignedArrayCount = UInt32(mutableColorsArray.count)
        var unsignedRandomNumber = arc4random_uniform(unsignedArrayCount)
        newColorIndexNumber = Int(unsignedRandomNumber)
    }
    currentColorIndexNumber = newColorIndexNumber
    return newColorIndexNumber
}

func randomColor() -> UIColor {
    var randomIndex = randomNumber()
    var randomColor = mutableColorsArray[randomIndex] as UIColor
    mutableColorsArray.removeAtIndex(randomIndex)
    if mutableColorsArray.count == 0 {
        mutableColorsArray = colorsArray
    }
    return randomColor
}

Answer 1

嘗試改組。 遵循以下原則：

var colorsArray = ["red","green","blue","yellow","purple"]
var currentColorIndexNumber = colorsArray.count

func shuffle<T>(inout array: Array<T>) -> Void {
  for index in 0..<colorsArray.count {
    let swapLocation = index + Int(arc4random_uniform(UInt32(colorsArray.count - index)))
    if index != swapLocation {
      swap(&array[swapLocation], &array[index])
    }
  }
}

func randomColor() -> String {
  // shuffle first time through, then every time the list is exhausted
  if currentColorIndexNumber >= colorsArray.count - 1 {
    // shuffle, but guarantee first color after shuffling
    // is not the same as last color from previous round
    do {
      shuffle(&colorsArray)
    } while colorsArray[0] == lastColor
    currentColorIndexNumber = -1
    lastColor = colorsArray[colorsArray.count-1]
  }
  currentColorIndexNumber += 1
  return colorsArray[currentColorIndexNumber]
}

for i in (5*colorsArray.count) {
  print("\(randomColor()) ")
  if i % colorsArray.count == 0 {
    println()
  }
}

Answer 2

我的成績很好

func randomObject(array: [AnyObject]) -> AnyObject {
    return array[Int(arc4random_uniform(UInt32(array.count)))]
}

要以隨機順序枚舉所有元素，只需使用下面的shuffle函數，按順序使用shuffled數組的元素，並在數組用盡后重新隨機排列。

func shuffleArray<T>(array: Array<T>) -> Array<T>  {
  for var i = array.count - 1; i > 0; i--  {
    var j = Int(arc4random_uniform(UInt32(i-1)))
    swap(&array[i], &array[j])
  }
  return array
}