[LeetCode] 681. Next Closest Time

gaomysion 發(fā)布于2019-08-16 13:34 / 1329人閱讀

摘要：找到所有可用的數(shù)，排序后存起來從右到左對進(jìn)行操作，只要有當(dāng)前最小單位時(shí)間的替換，返回替換后的時(shí)間找到的位置，然后加得到下一個(gè)位置如果下一個(gè)位置的數(shù)還是原來的數(shù)，或者超過了上限數(shù)，前進(jìn)到再下一個(gè)

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(shù)，排序后存起來
        char[] digits = new char[]{res[0], res[1], res[3], res[4]};
        Arrays.sort(digits);
        
        //從右到左對res進(jìn)行操作，只要有當(dāng)前最小單位時(shí)間的替換，返回替換后的時(shí)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，然后加1得到下一個(gè)位置
        int pos = Arrays.binarySearch(digits, cur)+1;
        //如果下一個(gè)位置的數(shù)還是原來的數(shù)，或者超過了上限數(shù)，前進(jìn)到再下一個(gè)
        while (pos < 4 && (digits[pos] == cur || digits[pos] > upper)) pos++;
        return pos == 4 ? digits[0] : digits[pos];
    }
}