小时转换的Go算法无法正确转换。

huangapple
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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？

  1. package main
  3. import (
  4.     "fmt"
  5.     "strconv"
  6.     "strings"
  7. )
  9. type Conversion struct {
  10.     conversion string
  11. }
  13. func timeConversion(s string) string {
  14.     var conversion Conversion
  15.     firstValue := fmt.Sprintf("%s%s", string(s[0]), string(s[1]))
  16.     secondValue := fmt.Sprintf("%s%s", string(s[3]), string(s[4]))
  17.     firstValueNumber, _ := strconv.ParseInt(firstValue, 0, 16)
  19.     fmt.Print()
  21.     if strings.Contains(s, "A") {
  22.         if firstValue == "12" {
  23.             conversion.conversion = fmt.Sprintf("%s:%s:%s%s", "00", secondValue, string(s[len(s)-4]), string(s[len(s)-3]))
  24.         } else {
  25.             conversion.conversion = fmt.Sprintf("%s:%s:%s%s", firstValue, secondValue, string(s[len(s)-4]), string(s[len(s)-3]))
  26.         }
  27.     } else if strings.Contains(s, "P") {
  28.         if firstValue == "12" {
  29.             conversion.conversion = fmt.Sprintf("%s:%s:%s%s", "12", secondValue, string(s[len(s)-4]), string(s[len(s)-3]))
  30.         } else if firstValue == "09" {
  31.             conversion.conversion = fmt.Sprintf("%s:%s:%s%s", "21", secondValue, string(s[len(s)-4]), string(s[len(s)-3]))
  32.         } else {
  33.             conversion.conversion = fmt.Sprintf("%d:%s:%s%s", firstValueNumber+12, secondValue, string(s[len(s)-4]), string(s[len(s)-3]))
  34.         }
  35.     }
  37.     return conversion.conversion
  38. }
  40. func main() {
  41.     fmt.Print(timeConversion("09:08:23PM"))
  42. }
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?

  1. package main
  2. import (
  3. "fmt"
  4. "strconv"
  5. "strings"
  6. )
  7. type Conversion struct {
  8. conversion string
  9. }
  10. func timeConversion(s string) string {
  11. var conversion Conversion
  12. firstValue := fmt.Sprintf("%s%s", string(s[0]), string(s[1]))
  13. secondValue := fmt.Sprintf("%s%s", string(s[3]), string(s[4]))
  14. firstValueNumber, _ := strconv.ParseInt(firstValue, 0, 16)
  15. fmt.Print()
  16. if strings.Contains(s, "A") {
  17. if firstValue == "12" {
  18. conversion.conversion = fmt.Sprintf("%s:%s:%s%s", "00", secondValue, string(s[len(s)-4]), string(s[len(s)-3]))
  19. } else {
  20. conversion.conversion = fmt.Sprintf("%s:%s:%s%s", firstValue, secondValue, string(s[len(s)-4]), string(s[len(s)-3]))
  21. }
  22. } else if strings.Contains(s, "P") {
  23. if firstValue == "12" {
  24. conversion.conversion = fmt.Sprintf("%s:%s:%s%s", "12", secondValue, string(s[len(s)-4]), string(s[len(s)-3]))
  25. } else if firstValue == "09" {
  26. conversion.conversion = fmt.Sprintf("%s:%s:%s%s", "21", secondValue, string(s[len(s)-4]), string(s[len(s)-3]))
  27. } else {
  28. conversion.conversion = fmt.Sprintf("%d:%s:%s%s", firstValueNumber+12, secondValue, string(s[len(s)-4]), string(s[len(s)-3]))
  29. }
  30. }
  31. return conversion.conversion
  32. }
  33. func main() {
  34. fmt.Print(timeConversion("09:08:23PM"))
  35. }

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

  1. firstValueNumber, err := strconv.ParseInt(firstValue, 0, 64)
  2. if err != nil {
  3.      fmt.Printf("%s", err.Error())
  4. }
  5. // "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

  1. firstValueNumber, err := strconv.ParseInt(firstValue, 0, 64)
  2. if err != nil {
  3. fmt.Printf("%s", err.Error())
  4. }
  5. // "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.

huangapple
  • 本文由 发表于 2023年3月31日 04:58:12
  • 转载请务必保留本文链接：https://go.coder-hub.com/75892938.html
  • algorithm
  • debugging
  • go
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