我正在寻找一个我可以调用的函数,它会告诉我所有(p)人同时可用的日期时间范围。 请帮我看看Objective-c或Swift语言。
p1: start: "2016-01-01 12:00", end: "2016-05-01 03:00"
p2: start: "2016-01-01 03:00", end: "2016-05-01 03:00"
p3: start: "2016-01-01 03:00", end: "2016-04-30 13:31"
在上面的例子中,答案应该是:
start: 2016-04-30 12:00, end: 2016-04-30 13:31
答案 0 :(得分:6)
将您的日期对转换为NSDateInterval对象并获取它们的交集:
文档甚至提供了一个相当不错的图表:
答案 1 :(得分:2)
您需要执行以下步骤:
这是一个我可以在swift中提出的快速代码:
func answer() {
let dateFormat = "yyyy-MM-dd HH:mm Z"
// Date ranges
let times = [["start": "2016-01-01 12:00 +0000", "end": "2016-05-01 03:00 +0000"],
["start": "2016-01-01 03:00 +0000", "end": "2016-05-01 03:00 +0000"],
["start": "2016-01-01 03:00 +0000", "end": "2016-04-30 13:31 +0000"]]
var intervals = [DateInterval]()
// Loop through date ranges to convert them to date intervals
for item in times {
if let start = convertStringToDate(string: item["start"]!, withFormat: dateFormat),
let end = convertStringToDate(string: item["end"]!, withFormat: dateFormat) {
intervals.append(DateInterval(start: start, end: end))
}
}
// Check for intersection
let intersection = intersect(intervals: intervals)
print(intersection)
}
// Converts the string to date with given format
func convertStringToDate(string: String, withFormat format: String) -> Date? {
let dateFormatter = DateFormatter()
dateFormatter.dateFormat = format
return dateFormatter.date(from: string)
}
// Cehck for intersection among the intervals in the given array and return
// the interval if found.
func intersect(intervals: [DateInterval]) -> DateInterval? {
// Algorithm:
// We will compare first two intervals.
// If an intersection is found, we will save the resultant interval
// and compare it with the next interval in the array.
// If no intersection is found at any iteration
// it means the intervals in the array are disjoint. Break the loop and return nil
// Otherwise return the last intersection.
var previous = intervals.first
for (index, element) in intervals.enumerated() {
if index == 0 {
continue
}
previous = previous?.intersection(with: element)
if previous == nil {
break
}
}
return previous
}
注意:请使用多个示例进行测试。我测试了以上日期范围及其工作正常。