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