# 681 Next Closest Time

## 681. [Next Closest Time](https://leetcode.com/problems/next-closest-time/description/)

## 1. Question

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.
```

## 2. Implementation

**(1) DFS (Enumeration)**

**思路:** 用**DFS**枚举所有可能的time，尝试对time的每个digit变为可用digit，比较谁离原来的time最近

```java
class Solution {
    int hour;
    int min;
    int diff = Integer.MAX_VALUE;

    public String nextClosestTime(String time) {
        if (time == null || time.length() == 0) {
            return time;
        }

        int[] digit = new int[4];
        String[] val = time.split(":");
        hour = Integer.parseInt(val[0]);
        min = Integer.parseInt(val[1]);

        digit[0] = hour / 10;
        digit[1] = hour % 10;
        digit[2] = min / 10;
        digit[3] = min % 10;

        String[] res = new String[1];

        getNextClosestTime(digit, 0, new int[4], res);
        return res[0];
    }

    public void getNextClosestTime(int[] digit, int index, int[] temp, String[] res) {
        if (index == 4) {
            int curHour = 10 * temp[0] + temp[1];
            int curMin = 10 * temp[2] + temp[3];
            if (curHour >= 0 && curHour <= 23 && curMin >= 0 && curMin <= 59) {
                int curDiff = getDiff(curHour, curMin);

                if (curDiff < diff) {
                    diff = curDiff;
                    res[0] = formatTime(curHour) + ":" + formatTime(curMin);
                }
            }
        }
        else {
            for (int i = 0; i < 4; i++) {
                temp[index] = digit[i];
                getNextClosestTime(digit, index + 1, temp, res);
            }
        }
    }

    public int getDiff(int curHour, int curMin) {
        int diff1 = 3600 - (60 * hour + min);
        int diff2 = 3600 - (60 * curHour + curMin);
        return diff2 < diff1 ? diff1 - diff2 : diff1 - diff2 + 3600;
    }

    public String formatTime(int time) {
        if (time >= 0 && time <= 9) {
            return "0" + time;
        }
        return "" + time;
    }
}
```

## 3. Time & Space Complexity

**DFS:** 时间复杂度O(1), 总共只有4 \* 4 \* 4 \* 4 = 256种情况, 空间复杂度O(1), 深度为time的digit的个数为4


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