[LeetCode] 681. Next Closest Time

Problem

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.

Solution

class Solution {
    public String nextClosestTime(String time) {
        char[] res = time.toCharArray();
        //找到所有可用的數，排序後存起來
        char[] digits = new char[]{res[0], res[1], res[3], res[4]};
        Arrays.sort(digits);
        
        //從右到左對res進行操作，只要有當前最小單位時間的替換，返回替換後的時間
        res[4] = findNext(digits, res[4], `9`);
        if (res[4] > time.charAt(4)) return String.valueOf(res);
        res[3] = findNext(digits, res[3], `5`);
        if (res[3] > time.charAt(3)) return String.valueOf(res);
        res[1] = res[0] == `2` ? findNext(digits, res[1], `3`) : findNext(digits, res[1], `9`);
        if (res[1] > time.charAt(1)) return String.valueOf(res);
        res[0] = findNext(digits, res[0], `2`);
        return String.valueOf(res);
    }
    private char findNext(char[] digits, char cur, char upper) {
        if (cur == upper) return digits[0];
        //找到cur的位置，然後加1得到下一個位置
        int pos = Arrays.binarySearch(digits, cur)+1;
        //如果下一個位置的數還是原來的數，或者超過了上限數，前進到再下一個
        while (pos < 4 && (digits[pos] == cur || digits[pos] > upper)) pos++;
        return pos == 4 ? digits[0] : digits[pos];
    }
}