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.

给一个时间，只能用原时间里的数字，求能组成的最近的下一个时间点，当下个时间点超过零点时，就当第二天的时间。

解法1：Brute fore，由于给的时间只到分钟，一天中有1440个分钟，也就是1440个时间点，可以从当前时间点开始，遍历一天的1440个时间点，每到一个时间点，看其所有的数字是否在原时间点字符中存在，如果不存在，直接break，然后开始遍历下一个时间点，如果四个数字都存在，说明找到了，换算成题目的时间形式返回即可。

解法2：替换数字。先把出现的数字去重排序，然后从最低位的分钟开始替换，如果低位分钟上的数字已经是最大的数字，就把低位分钟上的数字换成最小的数字，否则就将低位分钟上的数字换成下一个较大的数字。高位分钟上的数字已经是最大的数字，或则下一个数字大于5，那么直接换成最小值，否则就将高位分钟上的数字换成下一个较大的数字。小时数字也是同理，小时低位上的数字情况比较复杂，当小时高位不为2的时候，低位可以是任意数字，而当高位为2时，低位需要小于等于3。对于小时高位，其必须要小于等于2。Time: O(4*10)，Space: O(10)

解法3：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。Time: O(4^4)，Space: O(4)。

Java: 1

class Solution {

    public String nextClosestTime(String time) {

        int hour = Integer.parseInt(time.substring(0, 2));

        int min = Integer.parseInt(time.substring(3, 5));

        while (true) {

            if (++min == 60) {

                min = 0;

                ++hour;

                hour %= 24;

            }

            String curr = String.format("%02d:%02d", hour, min);

            Boolean valid = true;

            for (int i = 0; i < curr.length(); ++i)

                if (time.indexOf(curr.charAt(i)) < 0) {

                    valid = false;

                    break;

                }

            if (valid) return curr;

        }

    }

}

Java: 2

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);

    }

Java: 3

int diff = Integer.MAX_VALUE;

    String result = "";

    public String nextClosestTime(String time) {

        Set<Integer> 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<Integer> 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<Integer> 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);

        }

    }

Python:

class Solution(object):

    def nextClosestTime(self, time):

        """

        :type time: str

        :rtype: str

        """

        h, m = time.split(":")

        curr = int(h) * 60 + int(m)

        result = None

        for i in xrange(curr+1, curr+1441):

            t = i % 1440

            h, m = t // 60, t % 60

            result = "%02d:%02d" % (h, m)

            if set(result) <= set(time):

                break

        return result

Python:

class Solution(object):

    def nextClosestTime(self, time):

        """

        :type time: str

        :rtype: str

        """

        time = time[:2] + time[3:]

        isValid = lambda t: int(t[:2]) < 24 and int(t[2:]) < 60

        stime = sorted(time)

        for x in (3, 2, 1, 0):

            for y in stime:

                if y <= time[x]: continue

                ntime = time[:x] + y + (stime[0] * (3 - x))

                if isValid(ntime): return ntime[:2] + ':' + ntime[2:]

        return stime[0] * 2 + ':' + stime[0] * 2

C++:

class Solution {

public:

    string nextClosestTime(string time) {

        string res = "0000";

        vector<int> v{600, 60, 10, 1};

        int found = time.find(":");

        int cur = stoi(time.substr(0, found)) * 60 + stoi(time.substr(found + 1));

        for (int i = 1, d = 0; i <= 1440; ++i) {

            int next = (cur + i) % 1440;

            for (d = 0; d < 4; ++d) {

                res[d] = '0' + next / v[d];

                next %= v[d];

                if (time.find(res[d]) == string::npos) break;

            }

            if (d >= 4) break;

        }

        return res.substr(0, 2) + ":" + res.substr(2);

    }

};

C++:　　

class Solution {

public:

    string nextClosestTime(string time) {

        string res = time;

        set<int> s{time[0], time[1], time[3], time[4]};

        string str(s.begin(), s.end());

        for (int i = res.size() - 1; i >= 0; --i) {

            if (res[i] == ':') continue;

            int pos = str.find(res[i]);

            if (pos == str.size() - 1) {

                res[i] = str[0];

            } else {

                char next = str[pos + 1];

                if (i == 4) {

                    res[i] = next;

                    return res;

                } else if (i == 3 && next <= '5') {

                    res[i] = next;

                    return res;

                } else if (i == 1 && (res[0] != '2' || (res[0] == '2' && next <= '3'))) {

                    res[i] = next;

                    return res;

                } else if (i == 0 && next <= '2') {

                    res[i] = next;

                    return res;

                }

                res[i] = str[0];

            }

        }

        return res;

    }

};

All LeetCode Questions List 题目汇总

巴特西

[LeetCode] 681. Next Closest Time 下一个最近时间点

All LeetCode Questions List 题目汇总

最新文章

热门文章