class Solution {
public:
string maximumTime(string time) {
}
};
1736. 替换隐藏数字得到的最晚时间
给你一个字符串 time ，格式为 hh:mm（小时：分钟），其中某几位数字被隐藏（用 ? 表示）。
有效的时间为 00:00 到 23:59 之间的所有时间，包括 00:00 和 23:59 。
替换 time 中隐藏的数字，返回你可以得到的最晚有效时间。
示例 1：
输入：time = "2?:?0" 输出："23:50" 解释：以数字 '2' 开头的最晚一小时是 23 ，以 '0' 结尾的最晚一分钟是 50 。
示例 2：
输入：time = "0?:3?" 输出："09:39"
示例 3：
输入：time = "1?:22" 输出："19:22"
提示：
time 的格式为 hh:mm原站题解
python3 解法, 执行用时: 48 ms, 内存消耗: 16 MB, 提交时间: 2023-09-27 15:24:15
class Solution:
def maximumTime(self, time: str) -> str:
time = list(time)
if time[0] == '?':
time[0] = '2' if time[1] in ('0','1','2','3','?') else '1'
if time[1] == '?':
time[1] = '9' if time[0] in ('0','1') else '3'
if time[3] == '?': time[3] = '5'
if time[4] == '?': time[4] = '9'
return "".join(time)
python3 解法, 执行用时: 44 ms, 内存消耗: 15.9 MB, 提交时间: 2023-09-27 15:23:52
class Solution:
def maximumTime(self, time: str) -> str:
time = list(time)
for i in range(len(time)):
if time[i] == "?":
if i == 0:
time[i] = "2" if time[i+1] in "?0123" else "1"
elif i == 1:
time[i] = "3" if time[0] == "2" else "9"
elif i == 3:
time[i] = "5"
else:
time[i] = "9"
return "".join(time)
php 解法, 执行用时: 8 ms, 内存消耗: 18.5 MB, 提交时间: 2023-09-27 15:23:21
class Solution {
/**
* @param String $time
* @return String
*/
function maximumTime($time) {
//小时部分如果都是?，则取23点
//如果只有一个？，则取值会受另一个值的限制
if ($time[0] == '?' && $time[1] == '?') {
$time[0] = '2';
$time[1] = '3';
} elseif ($time[0] == '?') {
$time[0] = $time[1] > '3' ? '1' : '2';
} elseif ($time[1] == '?') {
$time[1] = $time[0] == '2' ? '3' : '9';
}
//分钟部分比较简单，直接取最大值即可
if ($time[3] == '?') $time[3] = '5';
if ($time[4] == '?') $time[4] = '9';
return $time;
}
}
javascript 解法, 执行用时: 64 ms, 内存消耗: 40.9 MB, 提交时间: 2023-09-27 15:21:30
/**
* @param {string} time
* @return {string}
*/
var maximumTime = function(time) {
const arr = Array.from(time);
if (arr[0] === '?') {
arr[0] = ('4' <= arr[1] && arr[1] <= '9') ? '1' : '2';
}
if (arr[1] === '?') {
arr[1] = (arr[0] == '2') ? '3' : '9';
}
if (arr[3] === '?') {
arr[3] = '5';
}
if (arr[4] === '?') {
arr[4] = '9';
}
return arr.join('');
};
java 解法, 执行用时: 0 ms, 内存消耗: 39.6 MB, 提交时间: 2023-09-27 15:21:15
class Solution {
public String maximumTime(String time) {
char[] arr = time.toCharArray();
if (arr[0] == '?') {
arr[0] = ('4' <= arr[1] && arr[1] <= '9') ? '1' : '2';
}
if (arr[1] == '?') {
arr[1] = (arr[0] == '2') ? '3' : '9';
}
if (arr[3] == '?') {
arr[3] = '5';
}
if (arr[4] == '?') {
arr[4] = '9';
}
return new String(arr);
}
}
cpp 解法, 执行用时: 4 ms, 内存消耗: 6.3 MB, 提交时间: 2023-09-27 15:21:00
class Solution {
public:
string maximumTime(string time) {
if (time[0] == '?') {
time[0] = ('4' <= time[1] && time[1] <= '9') ? '1' : '2';
}
if (time[1] == '?') {
time[1] = (time[0] == '2') ? '3' : '9';
}
if (time[3] == '?') {
time[3] = '5';
}
if (time[4] == '?') {
time[4] = '9';
}
return time;
}
};
golang 解法, 执行用时: 0 ms, 内存消耗: 1.9 MB, 提交时间: 2021-07-02 10:29:43
func maximumTime(time string) string {
arr := strings.Split(time, ":")
hour, minute := []byte(arr[0]), []byte(arr[1])
if hour[0] == '?' {
if hour[1] < '4' || hour[1] > '9' {
hour[0] = '2'
} else {
hour[0] = '1'
}
}
if hour[1] == '?' {
if hour[0] < '2' {
hour[1] = '9'
} else {
hour[1] = '3'
}
}
if minute[0] == '?' {
minute[0] = '5'
}
if minute[1] == '?' {
minute[1] = '9'
}
return string(hour) + ":" + string(minute)
}