ios - 如何在 iOS 中只获取一次随机数

Question

我想为 iOS 开发一款励志应用。使用此代码随机选择动机句子。

arc4random_uniform(100)

问题: 代码显示示例 2 和另一个 2，以便用户再次获得相同的句子。

我需要从某个范围(例如 100)中仅获取一次随机数的东西，并且在每个数字之后代码应该重复自身，以便用户可以再次看到激励性句子。

Answer 1

您可以使用 Set 来过滤已经提取的数字(Sets 包含唯一元素)，并执行类似这样的操作

import Foundation

var setOfNumbers: Set<UInt32> = []

func extractUniqueNumber(with treshold: UInt32, alreadyExtracted: inout Set<UInt32>) -> UInt32? {

    guard alreadyExtracted.count < treshold else { return nil }

    let randomNumber = arc4random_uniform(treshold)

    if !alreadyExtracted.contains(randomNumber) {
        alreadyExtracted.insert(randomNumber)
        return randomNumber
    } else {
        return extractUniqueNumber(with: treshold, alreadyExtracted: &alreadyExtracted)
    }
}

// test it out with a number of tries

for _ in 0...100 {
    if let number = extractUniqueNumber(with: 100, alreadyExtracted: &setOfNumbers) {
        print("Your number is: \(number).\n- Previously extracted numbers) are: \(setOfNumbers)")
    } else {
        print("You have extracted all  numbers already")
    }
}

如果提取数字的顺序很重要，可以将 Set 替换为 Array。付出的代价是对已经提取的数字的搜索速度变慢，但在 100 的范围内很可能仍然没问题。在这种情况下，代码将是

var setOfNumbers: [UInt32] = []

func extractUniqueNumber(with treshold: UInt32, alreadyExtracted: inout [UInt32]) -> UInt32? {

    guard alreadyExtracted.count < treshold else { return nil }

    let randomNumber = arc4random_uniform(treshold)

    if !alreadyExtracted.contains(randomNumber) {
        alreadyExtracted.append(randomNumber)
        return randomNumber
    } else {
        return extractUniqueNumber(with: treshold, alreadyExtracted: &alreadyExtracted)
    }
}