LeetCode Next Closest Time

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

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.

First get integer hash value of current time.

Get all 4 digits of current time. Take 2 of them to construct hour, and the other 2 to construct minute.

Use new time hash value - current time hash value and maintain the minimum diff.

If minused result < 0, it is next day, add it with 24 * 60.

Time Complexity: O(1). 共有4^4种组合.

Space: O(1). size为4的HashSet.

AC Java:

 class Solution {
     public String nextClosestTime(String time) {
         Set<Integer> hs = new HashSet<>();
         for(int i = 0; i<time.length(); i++){
             char c = time.charAt(i);
             if(c != ':'){
                 hs.add(c-'0');
             }
         }

         int cur = Integer.valueOf(time.substring(0, 2)) * 60 + Integer.valueOf(time.substring(3, 5)); 

         int minDiff = 24*60;
         // res is initialized as time, in case "11:11". The expected result is "11: 11".
         String res = time;
         for(int h1 : hs){
             for(int h2 : hs){
                 if(h1*10 + h2 < 24){
                     for(int m1 : hs){
                         for(int m2 : hs){
                             if(m1*10 + m2 < 60){
                                 int can = (h1*10+h2) * 60 + m1*10+m2;
                                 int diff = can-cur;
                                 if(diff < 0){
                                     diff += 24*60;
                                 }

                                 // diff can't 0, otherwise, it would return itself
                                 if(diff>0 && diff<minDiff){
                                     minDiff = diff;
                                     res = ""+h1+h2+":"+m1+m2;
                                     System.out.println("m1: "+ m1 + " m2: " + m2);
                                 }
                             }
                         }
                     }
                 }
             }
         }

         return res;
     }
 }