Medium
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.

看的答案，这种方法其实非常Straighforward, 就从给的time string每次加一分钟，然后检查这个新的time适不适合可以由原来的time所有的数字构成。如果可以，自然就是最近的了，因为我们是每次加一分钟这样加上来的。

这个方法有几个地方我不会，所以没想到：

• 一开始把string转换成int赋给hours, minutes, 方便后面的加分钟数；
• String.formart(“%02d:%02d”, hours, minutes)这种用法没有储备，%02d的意思就是保留两位整数，没有的位数补0，这样刚好就构成了hh:mm的格式，可以直接返回。这属于基础知识，但不常用，所以不知道。
• 最后检查curt是不是由time的数字组成的，用到了for循环curt的字符，用String.indexOf()看结果是不是<0来判断，非常方便，但我想不起这个方法，更不要说用它来解决算法题了，还是基础不扎实。

class Solution {
    //brute force
    //plus 1 minute everytime, check if this time can be constructed by the letter in the given string
    //if it can, return it because it's the closest else continue the adding process
    public String nextClosestTime(String time) {
        int hours = Integer.parseInt(time.substring(0,2));
        int mins = Integer.parseInt(time.substring(3,5));
        while (true){
            mins++;
            if (mins == 60){
                hours++;
                if (hours == 24){
                    hours =0;
                }
                mins = 0;
            }
            String curt = String.format("%02d:%02d", hours, mins);
            Boolean valid = true;
            for (int i = 0; i < curt.length(); i++){
                if (time.indexOf(curt.charAt(i)) < 0){
                    valid = false;
                    break;
                }
            }
            if (valid){
                return curt;
            }    
        }
    }
}