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.

 public class Test {
     public String nextClosestTime(String time) {
         char[] res = time.toCharArray();
         char[] digits = new char[] { res[], res[], res[], res[] };
         Arrays.sort(digits);

         // 从右到左对res进行操作，只要有当前最小单位时间的替换，返回替换后的时间
         res[] = findNext(digits, res[], '');
         if (res[] > time.charAt()) return String.valueOf(res);

         res[] = findNext(digits, res[], '');
         if (res[] > time.charAt()) return String.valueOf(res);

         res[] = res[] == '' ? findNext(digits, res[], '') : findNext(digits, res[], '');
         if (res[] > time.charAt()) return String.valueOf(res);

         res[] = findNext(digits, res[], '');
         return String.valueOf(res);
     }

     private char findNext(char[] digits, char cur, char upper) {
         if (cur == upper) return digits[];
         // 找到cur的位置，然后加1得到下一个位置
         int pos = Arrays.binarySearch(digits, cur) + ;
         // 如果下一个位置的数还是原来的数，或者超过了上限数，前进到再下一个
         while (pos <  && (digits[pos] == cur || digits[pos] > upper)) {
             pos++;
         }
         return pos ==  ? digits[] : digits[pos];
     }
 }