LeetCode Next Closest Time

Posted 2020-10-11 Dylan_Java_NYC

原题链接在这里：https://leetcode.com/problems/next-closest-time/description/

题目：

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.

题解：

用当前time已经出现的四个数字，组成新的time. 找diff最小的新time.

Time Complexity: O(1). 共有4^4种组合.

Space: O(1). size为4的HashSet.

AC Java:

 1 class Solution {
 2     public String nextClosestTime(String time) {
 3         if(time == null || time.length() != 5){
 4             throw new IllegalArgumentException("Invalid operation");
 5         }
 6         
 7         int cur = Integer.valueOf(time.substring(0,2)) * 60 + Integer.valueOf(time.substring(3));
 8         HashSet<Integer> hs = new HashSet<Integer>();
 9         for(int i = 0; i<time.length(); i++){
10             if(time.charAt(i) != ‘:‘){
11                 hs.add(time.charAt(i)-‘0‘);
12             }
13         }
14         
15         int minDiff = 24*60;
16         int res = cur;
17         for(int h1 : hs){
18             for(int h2 : hs){
19                 if(h1*10+h2 < 24){
20                     for(int m1 : hs){
21                         for(int m2 : hs){
22                             if(m1*10+m2 < 60){
23                                 int next = (h1*10+h2)*60 + (m1*10+m2);
24                                 int diff = Math.floorMod(next-cur, 24*60);
25                                 if(diff > 0 && diff < minDiff){
26                                     res = next;
27                                     minDiff = diff;
28                                 }
29                             }
30                         }
31                     }
32                 }
33             }
34         }
35         
36         return String.format("%02d:%02d", res/60, res%60);
37     }
38 }