# 问题 原题链接：[https://leetcode.com/problems/next-closest-time/](https://leetcode.com/problems/next-closest-time/) Given a time represented in the format "HH:MM", form the next closest time by reusing the current digits. There is no limit on how many times a digit can be reused. You may assume the given input string is always valid. For example, "01:34", "12:09" are all valid. "1:34", "12:9" are all invalid. Example 1: ``` Input: "19:34" Output: "19:39" Explanation: The next closest time choosing from digits 1, 9, 3, 4, is 19:39, which occurs 5 minutes later. It is not 19:33, because this occurs 23 hours and 59 minutes later. ``` Example 2: ``` Input: "23:59" Output: "22:22" Explanation: The next closest time choosing from digits 2, 3, 5, 9, is 22:22. It may be assumed that the returned time is next day's time since it is smaller than the input time numerically. ``` # 分析过程 - 输入：一个时间值 - 输出：不产生新的数字来组成下一个时间点 - 思路： - 一天有 1440 分钟，因此可以暴力扫描。把输入字符串拆分以后放入集合，根据输入的时间把整天的时间切成四段分别扫描。假设输入的是 12:34，四段分别扫描的区间为： - 12:35 ~ 12:59 - 13:00 ~ 23:59 - 00:00 ~ 11:59 - 12:00 ~ 12:35 - 按这样的顺序扫描，扫描时检查是不是所有的数字都在集合中，如果是就直接输出。 # 解决方法 ```swift class Solution { // 输入的字符串提取小时和分钟 func strToTime(timeStr: String) -> (Int, Int) { // 这里不检查合法性，假设输入的都是合法的 let strs = timeStr.components(separatedBy: ":") return (Int(strs[0])!, Int(strs[1])!) } // 检查整数的所有数字是不是包含在集合以内 func check(value: Int, inSet set: Set<Character>) -> Bool { for c in String.init(format: "%02d", value) { if !set.contains(c) { return false } } return true } func nextClosestTime(_ time: String) -> String { // 假设输入的都是合法的 // 数字提取出来放入集合 var digSet = Set<Character>() for c in time { if c != ":" { digSet.insert(c) } } // 取得输入时间的小时和分钟 let (h, m) = strToTime(timeStr: time) // 以下分成四种情况进行暴力扫描 // h:m ~ h:59 if m < 59 { for newM in (m + 1)...59 { if check(value: newM, inSet: digSet) { return String.init(format: "%02d:%02d", h, newM) } } } // h+1:00 ~ 23:59 if h < 23 { for newH in (h + 1)...23 { guard check(value: newH, inSet: digSet) else { continue } for newM in 0...59 { if check(value: newM, inSet: digSet) { return String.init(format: "%02d:%02d", newH, newM) } } } } // 00:00 ~ h-1:59 if h > 0 { for newH in 0...(h - 1) { guard check(value: newH, inSet: digSet) else { continue } for newM in 0...59 { if check(value: newM, inSet: digSet) { return String.init(format: "%02d:%02d", newH, newM) } } } } // h:00 ~ h:m for newM in 0...m { if check(value: newM, inSet: digSet) { return String.init(format: "%02d:%02d", h, newM) } } // 理论上不存在走到这里的可能 fatalError() } } ```