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75 lines
2.1 KiB
Markdown
75 lines
2.1 KiB
Markdown
# [1736. Latest Time by Replacing Hidden Digits](https://leetcode.com/problems/latest-time-by-replacing-hidden-digits/)
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## 题目
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You are given a string `time` in the form of `hh:mm`, where some of the digits in the string are hidden (represented by `?`).
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The valid times are those inclusively between `00:00` and `23:59`.
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Return *the latest valid time you can get from* `time` *by replacing the hidden* *digits*.
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**Example 1:**
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```
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Input: time = "2?:?0"
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Output: "23:50"
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Explanation: The latest hour beginning with the digit '2' is 23 and the latest minute ending with the digit '0' is 50.
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```
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**Example 2:**
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```
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Input: time = "0?:3?"
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Output: "09:39"
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```
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**Example 3:**
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```
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Input: time = "1?:22"
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Output: "19:22"
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```
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**Constraints:**
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- `time` is in the format `hh:mm`.
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- It is guaranteed that you can produce a valid time from the given string.
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## 题目大意
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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。
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## 解题思路
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- 简单题。根据题意，需要找到最晚的有效时间。枚举时间 4 个位置即可。如果第 3 个位置是 ？，那么它最晚时间是 5；如果第 4 个位置是 ？，那么它最晚时间是 9；如果第 2 个位置是 ？，那么它最晚时间是 9；如果第 1 个位置是 ？，根据第 2 个位置判断，如果第 2 个位置是大于 3 的数，那么第一个位置最晚时间是 1，如果第 2 个位置是小于 3 的数那么第一个位置最晚时间是 2 。按照上述规则即可还原最晚时间。
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## 代码
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```go
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package leetcode
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func maximumTime(time string) string {
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timeb := []byte(time)
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if timeb[3] == '?' {
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timeb[3] = '5'
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}
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if timeb[4] == '?' {
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timeb[4] = '9'
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}
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if timeb[0] == '?' {
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if int(timeb[1]-'0') > 3 && int(timeb[1]-'0') < 10 {
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timeb[0] = '1'
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} else {
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timeb[0] = '2'
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}
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}
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if timeb[1] == '?' {
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timeb[1] = '9'
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}
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if timeb[0] == '2' && timeb[1] == '9' {
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timeb[1] = '3'
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}
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return string(timeb)
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}
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``` |