Next Closest Time

　　　　这道题为中等题

　　题目:

　　　　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 r

　　思路:

　　　　个人思路: 这个题我首先把时间转换为字符串,而后循环获得c0,c1,c2,c3分别表示比当前位置数大的最小的一个数,首先比较最低位(好比上面第一个例子的4),若是有比他大的数那么就改变它的值再返回,而后再比较倒数第二位数,若是c1比他自己大且c1小于等于5,那么改变它的值并返回,同理再比较倒数第三个值,倒数第四个值,不然就所有返回最小值. 这个思路仍是比较清晰,可是有个缺点就是代码量太大了,若是遇见还有秒的话,就比较麻烦.

　　　　大神:保存现有集合,不断增长1分钟,当心，若是分钟超过59，增长小时，若是小时超过23，将小时重设为0,而后检查是否使用原始时间内的全部数字，以先到者为准.

　　代码:

　　　　个人代码:

 1 class Solution(object):
 2     def nextClosestTime(self, time):
 3         """
 4         :type time: str
 5         :rtype: str
 6         """
 7         a = [int(i) for i in time if i != ':']
 8         b = min(min(a[0], a[1]), min(a[2], a[3]))
 9         d = max(max(a[0], a[1]), max(a[2], a[3]))
10         c0 = d
11         c1 = d
12         c2 = d
13         c3 = d
14         
15         for i in a:
16             if i > a[0]: c0 = min(c0, i)
17             if i > a[1]: c1 = min(c1, i)
18             if i > a[2]: c2 = min(c2, i)
19             if i > a[3]: c3 = min(c3, i)
20         if a[3] < c3:
21             a[3] = c3
22         elif a[2] < c2 and c2 <= 5: 
23             a[2] = c2
24             a[3] = b
25         elif (a[1] < c1 and c1 <= 9 and a[0] <= 1) or (a[0] == 2 and a[1] < c1 and c1 <= 3):
26             a[1] = c1
27             a[2] = b
28             a[3] = b
29         elif a[0] < c0 and c0 <= 2:
30             a[0] = c0
31             a[1] = b
32             a[2] = b
33             a[3] = b
34         else:
35             a[0] = b
36             a[1] = b
37             a[2] = b
38             a[3] = b
39         a.insert(2,':')
40         return ''.join([str(i) for i in a])

　　　　大神代码:

　　　　

 1 class Solution(object):
 2     def nextClosestTime(self, time):
 3         """
 4         :type time: str
 5         :rtype: str
 6         """
 7         hour,minute = time.split(':')
 8         digits = set(map(int,[ digit for digit in hour + minute ]))
 9         while True:
10             hour, minute = int(hour), int(minute)
11             hour, minute = (hour, minute + 1) if minute < 59 else ((hour + 1)%24, 0)
12             hour = str(hour) if 10 <= hour < 24 else '0' + str(hour)
13             minute = str(minute) if 10 <= minute < 60 else '0' + str(minute)
14             current = set(map(int,[ digit for digit in hour + minute ]))
15             if not (current - digits): # current is proper subset of digits
16                 break
17         return hour + ':' + minute