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78 lines
1.9 KiB
Markdown
78 lines
1.9 KiB
Markdown
# [949. Largest Time for Given Digits](https://leetcode.com/problems/largest-time-for-given-digits/)
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## 题目
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Given an array of 4 digits, return the largest 24 hour time that can be made.
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The smallest 24 hour time is 00:00, and the largest is 23:59. Starting from 00:00, a time is larger if more time has elapsed since midnight.
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Return the answer as a string of length 5. If no valid time can be made, return an empty string.
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**Example 1**:
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```
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Input: [1,2,3,4]
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Output: "23:41"
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```
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**Example 2**:
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```
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Input: [5,5,5,5]
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Output: ""
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```
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**Note**:
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1. `A.length == 4`
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2. `0 <= A[i] <= 9`
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## 题目大意
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给定一个由 4 位数字组成的数组，返回可以设置的符合 24 小时制的最大时间。最小的 24 小时制时间是 00:00，而最大的是 23:59。从 00:00 （午夜）开始算起，过得越久，时间越大。以长度为 5 的字符串返回答案。如果不能确定有效时间，则返回空字符串。
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## 解题思路
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- 给出 4 个数字，要求返回一个字符串，代表由这 4 个数字能组成的最大 24 小时制的时间。
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- 简单题，这一题直接暴力枚举就可以了。依次检查给出的 4 个数字每个排列组合是否是时间合法的。例如检查 10 * A[i] + A[j] 是不是小于 24， 10 * A[k] + A[l] 是不是小于 60。如果合法且比目前存在的最大时间更大，就更新这个最大时间。
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## 代码
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```go
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package leetcode
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import "fmt"
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func largestTimeFromDigits(A []int) string {
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flag, res := false, 0
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for i := 0; i < 4; i++ {
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for j := 0; j < 4; j++ {
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if i == j {
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continue
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}
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for k := 0; k < 4; k++ {
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if i == k || j == k {
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continue
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}
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l := 6 - i - j - k
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hour := A[i]*10 + A[j]
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min := A[k]*10 + A[l]
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if hour < 24 && min < 60 {
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if hour*60+min >= res {
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res = hour*60 + min
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flag = true
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}
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}
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}
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}
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}
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if flag {
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return fmt.Sprintf("%02d:%02d", res/60, res%60)
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} else {
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return ""
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}
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}
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``` |