英文:
Go algorithm for hour convertion not converting correctly
问题
我遇到了一个HackerRank的挑战，需要编写一个函数，将给定格式的12小时制时间字符串转换为24小时制。
我使用下面的代码完成了这个任务，但是你可能注意到，对于下午09点这个特殊情况，我需要单独处理。每当我输入以09开头的小时数（只有这一个特殊情况），它会将09转换为12。所以我不得不创建一个特殊情况来处理它，虽然这样可以工作，但我想知道为什么会发生这种情况。你知道可能的问题是什么吗？
package main
import (
"fmt"
"strconv"
"strings"
)
type Conversion struct {
conversion string
}
func timeConversion(s string) string {
var conversion Conversion
firstValue := fmt.Sprintf("%s%s", string(s[0]), string(s[1]))
secondValue := fmt.Sprintf("%s%s", string(s[3]), string(s[4]))
firstValueNumber, _ := strconv.ParseInt(firstValue, 0, 16)
fmt.Print()
if strings.Contains(s, "A") {
if firstValue == "12" {
conversion.conversion = fmt.Sprintf("%s:%s:%s%s", "00", secondValue, string(s[len(s)-4]), string(s[len(s)-3]))
} else {
conversion.conversion = fmt.Sprintf("%s:%s:%s%s", firstValue, secondValue, string(s[len(s)-4]), string(s[len(s)-3]))
}
} else if strings.Contains(s, "P") {
if firstValue == "12" {
conversion.conversion = fmt.Sprintf("%s:%s:%s%s", "12", secondValue, string(s[len(s)-4]), string(s[len(s)-3]))
} else if firstValue == "09" {
conversion.conversion = fmt.Sprintf("%s:%s:%s%s", "21", secondValue, string(s[len(s)-4]), string(s[len(s)-3]))
} else {
conversion.conversion = fmt.Sprintf("%d:%s:%s%s", firstValueNumber+12, secondValue, string(s[len(s)-4]), string(s[len(s)-3]))
}
}
return conversion.conversion
}
func main() {
fmt.Print(timeConversion("09:08:23PM"))
}
英文:
I've come across a HackerRank challenge where I should build a function which takes a string in a given format of the time in the 12-hour format to the 24-hour format.
I managed to do that with the code below, but as you can notice, there's a special case for the 09 pm. Whenever I'd type in an hour starting with 09 (only this damn one) it'd convert the 09 to 12. So I had to actually create a specific case to treat that, which worked, but I'd like to understand why it happened. Do you know what the problem may be?
package main
import (
"fmt"
"strconv"
"strings"
)
type Conversion struct {
conversion string
}
func timeConversion(s string) string {
var conversion Conversion
firstValue := fmt.Sprintf("%s%s", string(s[0]), string(s[1]))
secondValue := fmt.Sprintf("%s%s", string(s[3]), string(s[4]))
firstValueNumber, _ := strconv.ParseInt(firstValue, 0, 16)
fmt.Print()
if strings.Contains(s, "A") {
if firstValue == "12" {
conversion.conversion = fmt.Sprintf("%s:%s:%s%s", "00", secondValue, string(s[len(s)-4]), string(s[len(s)-3]))
} else {
conversion.conversion = fmt.Sprintf("%s:%s:%s%s", firstValue, secondValue, string(s[len(s)-4]), string(s[len(s)-3]))
}
} else if strings.Contains(s, "P") {
if firstValue == "12" {
conversion.conversion = fmt.Sprintf("%s:%s:%s%s", "12", secondValue, string(s[len(s)-4]), string(s[len(s)-3]))
} else if firstValue == "09" {
conversion.conversion = fmt.Sprintf("%s:%s:%s%s", "21", secondValue, string(s[len(s)-4]), string(s[len(s)-3]))
} else {
conversion.conversion = fmt.Sprintf("%d:%s:%s%s", firstValueNumber+12, secondValue, string(s[len(s)-4]), string(s[len(s)-3]))
}
}
return conversion.conversion
}
func main() {
fmt.Print(timeConversion("09:08:23PM"))
}
答案1
得分: 2
如果在将字符串转换为整数的过程中捕获到错误，你会注意到对于大于7的值，转换会失败
firstValueNumber, err := strconv.ParseInt(firstValue, 0, 64)
if err != nil {
fmt.Printf("%s", err.Error())
}
// "08"导致无效语法firstValueNumber被打印
这是因为你将0作为基数传递给parseInt，告诉它从字符串的第一个字符推断基数。以0开头意味着八进制基数，因此大于7的数字将是无效的。改为传递10，你就不需要特殊处理了。
英文:
If you catch the error during your string to int conversion, you'll notice that for values >7, the conversion fails
firstValueNumber, err := strconv.ParseInt(firstValue, 0, 64)
if err != nil {
fmt.Printf("%s", err.Error())
}
// "08" results in invalid syntaxfirstValueNumber being printed
It happens because you pass 0 as the base, telling parseInt to infer the base from the first character of the string. A leading '0' implies octal base, and so digits greater than 7 would be invalid. Pass 10 instead and you won't need special handling.
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