这段代码从一个预设颜色数组中随机选择一种颜色。如何使得不会重复选择同一种颜色？
var colorArray = [(UIColor.redColor(), "red"), (UIColor.greenColor(), "green"), (UIColor.blueColor(), "blue"), (UIColor.yellowColor(), "yellow"), (UIColor.orangeColor(), "orange"), (UIColor.lightGrayColor(), "grey")]
var random = { () -> Int in
return Int(arc4random_uniform(UInt32(colorArray.count)))
} // makes random number, you can make it more reusable
var (sourceColor, sourceName) = (colorArray[random()])
var colorArray = [
(UIColor.redColor(), "red"),
(UIColor.greenColor(), "green"),
(UIColor.blueColor(), "blue"),
(UIColor.yellowColor(), "yellow"),
(UIColor.orangeColor(), "orange"),
(UIColor.lightGrayColor(), "grey")]
var indexes = [Int]();
func randomItem() -> UIColor
{
if indexes.count == 0
{
print("Filling indexes array")
indexes = Array(0..< colorArray.count)
}
let randomIndex = Int(arc4random_uniform(UInt32(indexes.count)))
let anIndex = indexes.removeAtIndex(randomIndex)
return colorArray[anIndex].0;
}
indexes。函数randomItem检查indexes是否为空。如果是，则使用从0到colorArray.count - 1的索引值填充它。indexes数组中选择一个随机索引，删除该索引处的值，并使用它从colorArray中获取和返回对象。(它不会从colorArray中删除对象。它使用间接方式，从最初包含每个条目的索引值的indexesArray中删除对象。return colorArray[anIndex].0。 - dimpiaxstride填充而不是范围。 - dimpiaxindexes = Array(0 ..< colorArray.count) 而不是使用 stride。 - Duncan Cimport Foundation // arc4random_uniform
class Random {
var r:UInt32
let max: UInt32
init(max: UInt32) {
self.max = max
r = arc4random_uniform(max)
}
var next: UInt32 {
var ret: UInt32
repeat {
ret = arc4random_uniform(max)
} while r == ret
r = ret
return r
}
}
// usage example
let r = Random(max: 5)
for i in 0..<10 {
print(r.r, r.next) // there will never be a pair of the same numbers in the
// generated stream
}
/*
2 4
4 0
0 3
3 0
0 3
3 4
4 1
1 3
3 4
4 3
*/
测试不同的K值和一百万流长度
class Random {
var r:UInt32
let max: UInt32
init(max: UInt32) {
self.max = max
r = arc4random_uniform(max)
}
var next: (UInt32, Int) {
var i = 0
var ret: UInt32
repeat {
ret = arc4random_uniform(max)
i += 1
} while r == ret
r = ret
return (r,i)
}
}
for k in 3..<16 {
let r = Random(max: UInt32(k))
var repetition = 0
var sum = 0
for i in 0..<1000000 {
let j = r.next
repetition = max(repetition, j.1)
sum += j.1
}
print("maximum of while repetition for k:", k, "is", repetition, "with average of", Double(sum) / Double(1000000) )
}
打印
maximum of while repetition for k: 3 is 15 with average of 1.499832
maximum of while repetition for k: 4 is 12 with average of 1.334008
maximum of while repetition for k: 5 is 9 with average of 1.250487
maximum of while repetition for k: 6 is 8 with average of 1.199631
maximum of while repetition for k: 7 is 8 with average of 1.167501
maximum of while repetition for k: 8 is 7 with average of 1.142799
maximum of while repetition for k: 9 is 8 with average of 1.124096
maximum of while repetition for k: 10 is 6 with average of 1.111178
maximum of while repetition for k: 11 is 7 with average of 1.099815
maximum of while repetition for k: 12 is 7 with average of 1.091041
maximum of while repetition for k: 13 is 6 with average of 1.083582
maximum of while repetition for k: 14 is 6 with average of 1.076595
maximum of while repetition for k: 15 is 6 with average of 1.071965
最后，这里有一个基于相同思路的更加Swifty和函数式的方法
import Foundation
func random(max: Int)->()->Int {
let max = UInt32(max)
var last = arc4random_uniform(max)
return {
var r = arc4random_uniform(max)
while r == last {
r = arc4random_uniform(max)
}
last = r
return Int(last)
}
}
let r0 = random(8)
let r1 = random(4)
for i in 0..<20 {
print(r0(), terminator: " ")
}
print("")
for i in 0..<20 {
print(r1(), terminator: " ")
}
/*
4 5 4 3 4 0 5 6 7 3 6 7 5 4 7 4 7 2 1 6
0 3 0 1 0 2 3 1 2 0 1 0 1 0 1 3 0 3 0 2
*/
0...8内生成一个随机位置，并将其插入到那里。var colorArray = [
(UIColor.redColor() , "red" ),
(UIColor.greenColor() , "green" ),
(UIColor.blueColor() , "blue" ),
(UIColor.yellowColor() , "yellow"),
(UIColor.orangeColor() , "orange"),
(UIColor.lightGrayColor(), "grey" )].shuffle() // shuffle() is from my link above
let spacing = 2 // Pick at least 2 colors before we see it again
if let randomColor = colorArray.popLast() {
colorArray.insert(randomColor,
atIndex: Int(arc4random_uniform(UInt32(colorArray.count - spacing))))
}
func getRandomItem<T>(arr: [T]) -> (unique: Bool) -> T {
var indexes: [Int]!
return { value in
let uniqIndex: Int
if value {
if indexes?.isEmpty != false {
indexes = [Int](0.stride(to: arr.count, by: 1))
}
uniqIndex = indexes.removeAtIndex(Int(arc4random_uniform(UInt32(indexes.count))))
}
else {
uniqIndex = Int(arc4random_uniform(UInt32(arr.count)))
}
return arr[uniqIndex]
}
}
let generate = getRandomItem(colorArray)
generate(unique: true).0 // greenColor
generate(unique: true).0 // redColor
generate(unique: true).0 // lightGrayColor
尝试一下，对我有效并且经过100%测试
let arrString = ["1","2","3","4","5","6"]
var selectedIndix = -1
@IBAction func btnClick(_ sender: Any) {
let randomElementIndex = randomElementString()
}
func randomElementString() -> Int{
let randomm = Int(arc4random_uniform(UInt32(arrString.count)))
if selectedIndix == randomm{
return randomElementString()
}else{
selectedIndix = randomm
return randomm
}
}
输出：
5121242316513126
尝试使用以下条件运行 while 循环：
while(self.source.backgroundColor == sourceColor) {
// get a new random sourceColor
}
这将一直循环，直到选择了一个新的随机颜色。
编辑
附加说明：重点是while循环。有办法防止无限循环，这取决于编程人员找到合适的解决方案。我认为SO不是一个地方来写别人的代码，而是提供建议..我的建议只是一个开始。
但由于我的答案得到了如此负面的评价，我将坚持而不是轻推朝着正确的方向前进。
其他答案过于臃肿。那又怎样？我上面提供的答案的时间复杂度不太理想。所以，这是我的新答案（伪代码）：
// array of all background colors
var arrayOfColors = [..]
// get a random index
var randomIndex = arc4random(size of arrayOfColors)
// select new background color
var newBGColor = arrayOfColors[randomIndex]
// old background color
var oldBGColor = self.source.backgroundColor
// remove new color from array (so that it's excluded from choices next time)
arrayOfColors.removeAtIndex(randomIndex)
// set the new color to the background
self.source.backgroundColor = newBGColor
// add current color back into the pool of potential colors
arrayOfColors.addObject(oldBGColor)