Question

代码首先在0-8之间生成随机数，并将其分配给var n。然后第二个randomNumber Generator函数循环n次以生成0到10之间的n个int，所有int都具有不同的发生概率并最终放入数组中。我想要的是没有那10个可能的数字重复，所以一旦选择了一个，它就不能被另一个n-1次运行func选择。我正在考虑循环重复或if语句或涉及索引的内容但我不确切知道如何，也不知道在括号内。谢谢你的帮助！有人低声说这是地球上最具挑战性和智能要求的编码难题。接受挑战？

import UIKit

let n = Int(arc4random_uniform(8))

var a:Double = 0.2
var b:Double = 0.3
var c:Double = 0.2
var d:Double = 0.3
var e:Double = 0.2
var f:Double = 0.1
var g:Double = 0.2
var h:Double = 0.4
var i:Double = 0.2
var j:Double = 0.2
var k: [Int] = []

for _ in 0...n {
    func randomNumber(probabilities: [Double]) -> Int {
        let sum = probabilities.reduce(0, +)
        let rnd = sum * Double(arc4random_uniform(UInt32.max)) / Double(UInt32.max)
        var accum = 0.0
        for (i, p) in probabilities.enumerated() {
            accum += p
            if rnd < accum {
                return i
            }}
        return (probabilities.count - 1)
    }
    k.append(randomNumber(probabilities: [a, b, c, d, e, f, g, h, i, j]))
}
print(k)

Answer 1

伪代码 -

1)generate a number between 1-8
    n
2)take empty array
    arr[]
3)loop from 0 to n
    1) generate a random no
        temp
    2) check if it is there in arr
            > if it is there in arr, generate another
    3) when you get a number which is not there in arr, insert it

这是一个python代码

import random

n = random.randint(1,8)
arr = []
print(n)
for each in range(n):
    temp = random.randint(1, 10)
    while temp in arr:
        temp = random.randint(1, 10)
        print(temp)
    arr.append(temp)
print(arr)

检查代码示例the docs

Answer 2

快速版Ankush的回答 -

 let n = arc4random_uniform(7) + 1
 var arr: [UInt32] = []
 for _ in 0 ... n {
    var temp = arc4random_uniform(9) + 1
    while arr.contains(temp) {
        temp = arc4random_uniform(9) + 1
    }
    print(temp)
    arr.append(temp)
 }
print(arr)

希望这有帮助！

生成随机计数的随机Int数组，无需重复变量

2 个答案: