Google
681 Next Closest Time

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:
1
Input: "19:34"
2
3
Output: "19:39"
4
5
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
6
, because this occurs 23 hours and 59 minutes later.
Copied!
Example 2:
1
Input: "23:59"
2
3
Output: "22:22"
4
5
Explanation: The next closest time choosing from digits 2, 3, 5, 9, is 22:22
6
. It may be assumed that the returned time is next day's time since it is smaller than the input time numerically.
Copied!

2. Implementation

(1) DFS (Enumeration)
思路: 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
1
class Solution {
2
int hour;
3
int min;
4
int diff = Integer.MAX_VALUE;
5
6
public String nextClosestTime(String time) {
7
if (time == null || time.length() == 0) {
8
return time;
9
}
10
11
int[] digit = new int[4];
12
String[] val = time.split(":");
13
hour = Integer.parseInt(val[0]);
14
min = Integer.parseInt(val[1]);
15
16
digit[0] = hour / 10;
17
digit[1] = hour % 10;
18
digit[2] = min / 10;
19
digit[3] = min % 10;
20
21
String[] res = new String[1];
22
23
getNextClosestTime(digit, 0, new int[4], res);
24
return res[0];
25
}
26
27
public void getNextClosestTime(int[] digit, int index, int[] temp, String[] res) {
28
if (index == 4) {
29
int curHour = 10 * temp[0] + temp[1];
30
int curMin = 10 * temp[2] + temp[3];
31
if (curHour >= 0 && curHour <= 23 && curMin >= 0 && curMin <= 59) {
32
int curDiff = getDiff(curHour, curMin);
33
34
if (curDiff < diff) {
35
diff = curDiff;
36
res[0] = formatTime(curHour) + ":" + formatTime(curMin);
37
}
38
}
39
}
40
else {
41
for (int i = 0; i < 4; i++) {
42
temp[index] = digit[i];
43
getNextClosestTime(digit, index + 1, temp, res);
44
}
45
}
46
}
47
48
public int getDiff(int curHour, int curMin) {
49
int diff1 = 3600 - (60 * hour + min);
50
int diff2 = 3600 - (60 * curHour + curMin);
51
return diff2 < diff1 ? diff1 - diff2 : diff1 - diff2 + 3600;
52
}
53
54
public String formatTime(int time) {
55
if (time >= 0 && time <= 9) {
56
return "0" + time;
57
}
58
return "" + time;
59
}
60
}
Copied!

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