下面是一个可能的实现,如何计算分数的十进制扩展并检测重复小数(即十进制扩展中的句点).它是在Code Review到Swift 的有理数的十进制扩展中发布和审查的代码的翻译.
该算法只是Long Division,而array(remainders)用于检查周期性的十进制扩展.(为简单起见,假设分子是非负的,分母是正的.如果需要,这可以推广.)
struct DecimalFraction : Printable {
let wholePart : Int // Integer part
let fractionDigits : [Int] // Fractional digits
let repeatingAt : Int? // Position of first repeating digit, or `nil`
// Create DecimalFraction from given fraction
init(numerator : Int, denominator : Int) {
precondition(numerator >= 0, "`numerator` must be non-negative")
precondition(denominator > 0, "`denominator` must be positive")
wholePart = numerator / denominator
var fractionDigits : [Int] = []
var repeatingAt : Int? = nil
var rem = (abs(numerator) % denominator) * 10
var remainders : [Int] = []
while (rem > 0 && repeatingAt == nil) {
remainders.append(rem)
let digit = rem / denominator
rem = (rem % denominator) * 10
fractionDigits.append(digit)
repeatingAt = find(remainders, rem)
}
self.fractionDigits = fractionDigits
self.repeatingAt = repeatingAt
}
// Produce a string description, e.g. "12.3{45}"
var description : String {
var result = String(wholePart) + "."
for (idx, digit) in enumerate(fractionDigits) {
if idx == repeatingAt {
result += "{"
}
result += String(digit)
}
if repeatingAt != nil {
result += "}"
}
return result
}
}
Run Code Online (Sandbox Code Playgroud)
例子:
println(DecimalFraction(numerator: 3, denominator: 8))
// 0.375
println(DecimalFraction(numerator: 1, denominator: 3))
// 0.{3}
println(DecimalFraction(numerator: 20, denominator: 7))
// 2.{857142}
println(DecimalFraction(numerator: 12222, denominator: 990))
// 12.3{45}
Run Code Online (Sandbox Code Playgroud)
句点简单地用花括号表示,但是应该很容易修改代码以产生一个NSAttributedString
表示句点的例子 - 例如 - 水平线.
| 归档时间: |
|
| 查看次数: |
859 次 |
| 最近记录: |