Next Closest Time问题及解法

问题描述：

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.

示例:

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.

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.

问题分析：

将time中的数字按升序存放的数组res中，分别对time的每一个数字time[i](i = 4,3,1,0)从后往前与res中的数进行比较，在res中找到第一个比time[i]中数字大的数s，将time[i] = s,i 之后的所有time中的数字，都置为res[0].

过程详见代码：

class Solution {
public:
string nextClosestTime(string time) {
vector<int> res{ time[0] - '0', time[1] - '0', time[3] - '0', time[4] - '0'};
sort(res.begin(), res.end());
for (int i = 0; i < 4; i++)
{
if (res[i] > time[4] - '0')
{
time[4] = res[i] + '0';
return time;
}
}
for (int i = 0; i < 4; i++)
{
if (res[i] > time[3] - '0' && res[i] < 6)
{
time[3] = res[i] + '0';
time[4] = res[0] + '0';
return time;
}
}
for (int i = 0; i < 4; i++)
{
if (res[i] > time[1] - '0' && (time[0] - '0' < 2 || res[i] < 4))
{
time[1] = res[i] + '0';
time[3] = res[0] + '0';
time[4] = res[0] + '0';
return time;

}
}
for (int i = 0; i < 4; i++)
{
if (res[i] > time[0] - '0' && res[i] < 3)
{
time[0] = res[i] + '0';
time[1] = res[0] + '0';
time[3] = res[0] + '0';
time[4] = res[0] + '0';
return time;
}
}
time[0] = res[0] + '0';
time[1] = res[0] + '0';
time[3] = res[0] + '0';
time[4] = res[0] + '0';
return time;
}
};