Question

在我的代码中，checkTheNextTime()函数的数组包含字符串00.00到23.59。通过编写此功能，我想找到最近的未来时间。但是当我尝试使用timeTable（在代码中显示）时，它返回23.30而不是23.32（现在是22.24）。我猜编译器从右到左搜索数组。我怎样才能找到最近的未来时间？

var timeTable = ["09.00","10.20","10.35","11.55","12.00","12.40","13.20","14.40","14.50", "23.00", "23.30", "23.31", "23.32"]
func checkTheNextTime(array array: Array<String>) -> String{


    var nextTime: String?
    for index in array {
        let generatedString:String = getTimeAsMinToCheck(finalTime: index)
        let indexInt = Int(generatedString)
        if indexInt > 0{
            nextTime = index
        }

    }

    return nextTime!
}


func getTimeAsMinToCheck(finalTime finalTime: String) -> String{

        let date = NSDate()
        let formatter = NSDateFormatter()
        formatter.timeStyle = NSDateFormatterStyle.NoStyle
        formatter.dateStyle = NSDateFormatterStyle.ShortStyle
        let now = formatter.stringFromDate(date)
        formatter.locale = NSLocale.systemLocale()
        formatter.dateFormat = "M/dd/yy HH.mm"
        let datetofinish = formatter.dateFromString("\(now) \(finalTime)")
        let finishDate: NSDate = datetofinish!
        let secondsFromNowToFinish = finishDate.timeIntervalSinceNow
        let minutes = Int(secondsFromNowToFinish / 60)
        return String(minutes)
    }

Answer 1

此代码应符合您的要求：完成 Swift 2.0 ：

    var timeTable = ["09.00","10.20","10.35","11.55","12.00","12.40","13.20","14.40","14.50", "23.00", "23.30", "23.31", "23.32"]

    func checkTheNextTime(array array: Array<String>) -> String{ 

        let currentTime:String = getTimeAsMinToCheck(finalTime: "23.24") // set this value by calculating from current time
        let currentTimeInt = Int(currentTime)// Int value of currentTime

        var nextTime: String? //this will hold the nearest future value

        var minDiff: Int = 24*60 //lets start with maximum value

        for index in timeTable {

            let generatedString:String = getTimeAsMinToCheck(finalTime: index)
            let indexInt = Int(generatedString)

            if (indexInt > currentTimeInt) { //checking for future time only

                let timeDiff = indexInt - currentTimeInt // this will be positive

                if (timeDiff < minDiff) {

                    minDiff = timeDiff //update minDiff as timeDiff is less than minDiff
                    nextTime = index
                }
            }
        }

        return nextTime!
    }

Answer 2

假设23是正确的答案（从上面的评论中不清楚），这里有一个使用swift 2.0和闭包的解决方案

将timeTable数组映射到当前的delta数组中时间（无效条目映射到0）
将最小增量添加到现在的时间

let timeNow: Float = 22.24
let timeTable = ["09.00","10.20","10.35","11.55","12.00","12.40","13.20","14.40","14.50", "23.00", "23.30", "23.31", "23.32"]

let minDelta = timeTable
  .map { Float(NSNumberFormatter().numberFromString($0) ?? 0.0) - timeNow }
  .filter { $0 > 0 }
  .minElement()
let nextTime = (minDelta ?? 0) + timeNow

print(nextTime)  // 23.0

如何搜索数组中的最小正值？

2 个答案: