### 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 returned time is next day's time since it is smaller than the input time numerically.


### 启发式解法

1. if the first digit is 0 or 1: Max = 19:59
2. if the first digit is 2: Max = 23:59

#### 代码

public String nextClosestTime(String time) {
char[] charTime = time.toCharArray();   // [1, 9, :, 3, 4]
char[] nums = Arrays.copyOf(charTime,charTime.length);
Arrays.sort(nums);                      // [1, 3, 4, 9, :]
/** the carry */
int i = 0;
outFor:
for (i = 4; i >= 0; i--) {
if (i == 2) { continue; } // skip ":" in the middle
char c = charTime[i];
int next = 0;
while (next < 4 && nums[next] <= c) { next++; }
// maximum legal value
char max = '9';
switch (i) {
case 3: max = '5'; break;
case 1: if (charTime == '2') { max = '4'; } break;
case 0: max = '2'; break;
}
if (next == 4 || nums[next] > max) { continue outFor; }
charTime[i] = nums[next];
break;
}
/** fill the remainder with smallest digit */
for (int j = i + 1; j <= 4; j++) {
if (j == 2) { continue; }
charTime[j] = nums; // at least one of [0,1,2]
}
return new String(charTime);
}


#### 结果 