如何在Swift中生成一个随机数而不重复以前的随机数？

Question

我对Swift和编程逻辑很陌生，所以请多多包涵

如何在Swift中生成0到9之间的随机数而不重复最后生成的数字？因为相同的数字不会连续出现两次。

Answer 1

Swift 5更新这是一个很好的技巧，可以从以前不仅仅选择的数字中进行平均选择。您有10个数字，但只想从9个数字中选择（0到9，但不包括前一个数字）。如果将范围减少1，则可以从9个随机数中进行选择，然后将重复的数字替换为该范围的前一个最高数字。这样，您每次只需生成一个随机数即可获得统一性。这可以Int.random(in:excluding:)在传递所需值的位置实现exclude。extension Int {&nbsp; &nbsp; static func random(in range: ClosedRange<Int>, excluding x: Int) -> Int {&nbsp; &nbsp; &nbsp; &nbsp; if range.contains(x) {&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; let r = Int.random(in: Range(uncheckedBounds: (range.lowerBound, range.upperBound)))&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; return r == x ? range.upperBound : r&nbsp; &nbsp; &nbsp; &nbsp; } else {&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; return Int.random(in: range)&nbsp; &nbsp; &nbsp; &nbsp; }&nbsp; &nbsp; }}例：// Generate 30 numbers in the range 1...3 without repeating the// previous number&nbsp;&nbsp;var r = Int.random(in: 1...3)for _ in 1...30 {&nbsp; &nbsp; r = Int.random(in: 1...3, excluding: r)&nbsp; &nbsp; print(r, terminator: " ")}print()1 3 2 1 2 1 3 2 1 3 1 3 2 3 1 2 3 2 1 3 2 1 3 1 2 3 2 1 2 1 3 2 3 2 1 3 1 2 1 2上一个答案：var previousNumber = arc4random_uniform(10)&nbsp; &nbsp;// seed the previous numberfunc randomNumber() -> UInt32 {&nbsp; &nbsp; var randomNumber = arc4random_uniform(9)&nbsp; // generate 0...8&nbsp; &nbsp; if randomNumber == previousNumber {&nbsp; &nbsp; &nbsp; &nbsp; randomNumber = 9&nbsp; &nbsp; }&nbsp; &nbsp; previousNumber = randomNumber&nbsp; &nbsp; return randomNumber}