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();
//找到所有可用的数,排序后存起来
char[] digits = new char[]{res[0], res[1], res[3], res[4]};
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[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,然后加1得到下一个位置
int pos = Arrays.binarySearch(digits, cur)+1;
//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 < 4 && (digits[pos] == cur || digits[pos] > upper)) pos++;
return pos == 4 ? digits[0] : digits[pos];
}
}
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