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.

一刷
題解：
對于"HH:M_", "HH:M", "H:MM", "_H:MM", apply 尋找下一個(gè)最大的digit.

1. 首先對所有的digit排序 "17:38" -> digits[] {'1', '3', '7', '8'}
2. 對于"HH:M_"，如果已經(jīng)是最大一位digits[3]， 比如8，那么下一個(gè)就為digits[0]
3. 對于"HH:M_", upperLimit是5(00-59), 那么我們就在(digits[2], 6)中尋找朝巫，如果找不到，那么就為digits[0]
4. 對于"H_:MM", 如果input[0] == 2, 那么upperlimit為2瞒斩，否則為9
5. 對于"_H:MM", upperlimit是2

findnext函數(shù)的思路是冯袍，我們在當(dāng)前的digit和upperlimit中尋找第一個(gè)大于當(dāng)前的digit，如果找不到稠氮，就選digit[0]

class Solution {
    
    public String nextClosestTime(String time) {
        char[] result = time.toCharArray();
        char[] digits = new char[] {result[0], result[1], result[3], result[4]};
        Arrays.sort(digits);
        
        // find next digit for HH:M_
        result[4] = findNext(result[4], (char)('9' + 1), digits);  // no upperLimit for this digit, i.e. 0-9
        if(result[4] > time.charAt(4)) return String.valueOf(result);  // e.g. 23:43 -> 23:44
        
        // find next digit for HH:_M
        result[3] = findNext(result[3], '5', digits);
        if(result[3] > time.charAt(3)) return String.valueOf(result);  // e.g. 14:29 -> 14:41
        
        // find next digit for H_:MM
        result[1] = result[0] == '2' ? findNext(result[1], '3', digits) : findNext(result[1], (char)('9' + 1), digits);
        if(result[1] > time.charAt(1)) return String.valueOf(result);  // e.g. 02:37 -> 03:00 
        
        // find next digit for _H:MM
        result[0] = findNext(result[0], '2', digits);
        return String.valueOf(result);  // e.g. 19:59 -> 11:11
    }
    
    /** 
     * find the next bigger digit which is no more than upperLimit. 
     * If no such digit exists in digits[], return the minimum one i.e. digits[0]
     * @param current the current digit
     * @param upperLimit the maximum possible value for current digit
     * @param digits[] the sorted digits array
     * @return 
     */
    private char findNext(char current, char upperLimit, char[] digits) {
        //System.out.println(current);
        if(current == upperLimit) {
            return digits[0];
        }
        int pos = Arrays.binarySearch(digits, current) + 1;
        while(pos < 4 && (digits[pos] > upperLimit || digits[pos] == current)) { // traverse one by one to find next greater digit
            pos++;
        }
        return pos == 4 ? digits[0] : digits[pos];
    }
    
}