Link: https://leetcode.cn/problems/next-closest-time/
Question
difficulty: mid
adj diff: 4
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: time = "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: time = "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.
Constraints:
time.length == 5
time is a valid time in the form "HH:MM".
0 <= HH < 24
0 <= MM < 60
Code
class Solution {
public String nextClosestTime(String time) {
List<String> possibleTimes = new ArrayList<>(); // 存储所有合理的事件
Set<Character> set = new HashSet<>();
List<String> hourList = new LinkedList<>(); // 存储合理的小时部分
List<String> monthList = new LinkedList<>(); // 存储合理的分钟部分
for(char digit: time.toCharArray()){
if(digit != ':') {
set.add(digit);
}
}
for(char ch1: set){
for(char ch2: set){
String str = ch1 + "" + ch2;
if (str.compareTo("24") < 0) {
hourList.add(str);
}
if (str.compareTo("60") < 0) {
monthList.add(str);
}
}
}
// 拼接小时与分钟部分
for(String h : hourList){
for(String m : monthList){
possibleTimes.add(h + ":" +m);
}
}
Collections.sort(possibleTimes); // 排序
int i = possibleTimes.indexOf(time); // 获取当前索引
if (i == possibleTimes.size() - 1) {
return possibleTimes.get(0);
} else {
return possibleTimes.get(i + 1);
}
}
}