681 Next Closest Time

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.

最优解法，时间复杂度为O(4*10)，空间复杂度为O(10)，思路：用一个伪哈希表来存给的时间的数字，然后从第四位开始，去哈希表里找有没有比给定的第四位大而且合法的数字，如果有则返回带有这个合法数字的新的string；如果没有，就把第四位设为哈希表里的最小值。依次类推，从第四位走到第一位，如果第一位还不存在一个比它大的合法值，则四位都设为最小值，然后返回，此时对应的情况就是example2里的情况，返回的是第二天的最小时间。此解法简单易懂，而且效率很高，beats 96%的leetcoder

public class NextClosestTime {
    public String nextClosetTime(String time) {
        char[] t = time.toCharArray(), result = new char[4];
        int[] list = new int[10];
        char min = '9';
        
        for (char c : t) {
            if (c == ':') {
                continue;
            }
            list[c - '0']++;
            if (c < min) {
                min = c;
            }
        }
        
        for (int i = t[4] - '0' + 1; i <= 9; i++) {
            if (list[i] != 0) {
                t[4] = (char)(i + '0');
                return new String(t);
            }
        }
        t[4] = min;
        
        for (int i = t[3] - '0' + 1; i <= 5; i++) {
            if (list[i] != 0) {
                t[3] = (char)(i + '0');
                return new String(t);
            }
        }
        t[3] = min;
        
        int stop = t[0] < '2' ? 9 : 3;
        for (int i = t[1] - '0' + 1; i <= stop; i++) {
            if (list[i] != 0) {
                t[1] = (char)(i + '0');
                return new String(t);
            }
        }
        t[1] = min;
        
        for (int i = t[0] - '0' + 1; i <= 2; i++) {
            if (list[i] != 0) {
                t[0] = (char)(i + '0');
                return new String(t);
            }
        }
        t[0] = min;
        return new String(t);
    }
    
    public static void main(String[] args) {
        NextClosestTime ts = new NextClosestTime();
        String time = "23:56";
        String res = ts.nextClosetTime(time);
        System.out.println(res);
    }
}