LeetCode 681. Next Closest Time（java）

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.

解法一：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 String nextClosestTime(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);
    }

解法二：找出这四个数字的所有可能的时间组合，然后和给定时间比较，maintain一个差值最小的，返回这个string。时间复杂度O(4^4)，空间复杂度O(4)。

int diff = Integer.MAX_VALUE;
    String result = "";

    public String nextClosestTime(String time) {
        Set set = new HashSet<>();
        set.add(Integer.parseInt(time.substring(0, 1)));
        set.add(Integer.parseInt(time.substring(1, 2)));
        set.add(Integer.parseInt(time.substring(3, 4)));
        set.add(Integer.parseInt(time.substring(4, 5)));

        if (set.size() == 1) return time;

        List digits = new ArrayList<>(set);
        int minute = Integer.parseInt(time.substring(0, 2)) * 60 + Integer.parseInt(time.substring(3, 5));

        dfs(digits, "", 0, minute);

        return result;
    }

    private void dfs(List digits, String cur, int pos, int target) {
        if (pos == 4) {
            int m = Integer.parseInt(cur.substring(0, 2)) * 60 + Integer.parseInt(cur.substring(2, 4));
            if (m == target) return;
            int d = m - target > 0 ? m - target : 1440 + m - target;
            if (d < diff) {
                diff = d;
                result = cur.substring(0, 2) + ":" + cur.substring(2, 4);
            }
            return;
        }

        for (int i = 0; i < digits.size(); i++) {
            if (pos == 0 && digits.get(i) > 2) continue;
            if (pos == 1 && Integer.parseInt(cur) * 10 + digits.get(i) > 23) continue;
            if (pos == 2 && digits.get(i) > 5) continue;
            if (pos == 3 && Integer.parseInt(cur.substring(2)) * 10 + digits.get(i) > 59) continue;
            dfs(digits, cur + digits.get(i), pos + 1, target);
        }
    }