[英]Convert Letters & Numbers in a Phone Number to all Numbers (Java)
import java.util.Scanner;
import javax.swing.JOptionPane;
public class PhonePadTranslator {
private static Scanner input;
public static void main(String[] args) {
input = new Scanner(System.in);
System.out.println("Enter The Phone Number (With Letters) ");
String initial_phone_number = input.nextLine();
initial_phone_number = initial_phone_number.toUpperCase();
int phone_number_final = 0;
System.out.printf("The phone number for %s is %s", initial_phone_number, phone_number_final);
}//end of main
public static int full_number(String initial_phone_number)
{
int which_character = 0;
int phone_number_final = 0;
char ch = (Character) null;
for (which_character = 0; which_character < initial_phone_number.length(); which_character++)
{
if (Character.isLetter(ch))
{
switch(ch)
{
case 'A' : case 'B' : case 'C' : phone_number_final = 2; break;
case 'D' : case 'E' : case 'F' : phone_number_final = 3; break;
case 'G' : case 'H' : case 'I' : phone_number_final = 4; break;
case 'J' : case 'K' : case 'L' : phone_number_final = 5; break;
case 'M' : case 'N' : case 'O' : phone_number_final = 6; break;
case 'P' : case 'Q' : case 'R' : case 'S' : phone_number_final = 7; break;
case 'T' : case 'U' : case 'V' : phone_number_final = 8; break;
case 'W' : case 'X' : case 'Y' : case 'Z' : phone_number_final =9; break;
}
return (char)phone_number_final;
}
if (Character.isDigit(ch))
{
return (char)phone_number_final;
}
else {
return (char)phone_number_final;
}
} //end of for
return ch;
}//end of full_number
}//end of class
我只是想我會復制/粘貼整個內容......但是每當我運行代碼時,它都會不斷輸出The phone number for 1800FLOWERS is 0 。 現在我確定還有其他一些錯誤,但我主要關心的是為什么它一直給我一個0
? 我覺得這是因為我將其初始化為該值,並且出於某種原因我從未更改過該值。 請幫忙,我的教授需要永遠回復我的電子郵件:(
改變
int phone_number_final = 0;
到
int phone_number_final = full_number(initial_phone_number);
您沒有將結果分配給您的變量。
除此之外,我相信您的full_number
函數也不完全正確。
更新代碼:
import java.util.Scanner;
public class StringToNumbers
{
private static Scanner input;
public static void main(String[] args)
{
input = new Scanner(System.in);
System.out.println("Enter The Phone Number (With Letters): ");
String initial_phone_number = input.nextLine();
initial_phone_number = initial_phone_number.toUpperCase();
long phone_number_final = full_number(initial_phone_number);
System.out.printf("%nOutput phone number for '%s' is '%s'",
initial_phone_number, phone_number_final);
}
public static long full_number(String initial_phone_number)
{
// Use long instead of int for 'number' if the string will be longer than max int value
// 2147483647, which is '10 digits'
long number = 0;
int strLen = initial_phone_number.length();
for (int currCharacter = 0; currCharacter < strLen; currCharacter++)
{
char ch = initial_phone_number.charAt(currCharacter);
// For A-Z & 0-9, multiply by 10, add the 'char' to number.
// i.e., Shift existing value to the left by 1 digit, add current 'char' to it
// Use long instead of int if the string will be longer than max int value (2147483647)
if (Character.isLetter(ch))
{
switch(ch)
{
case 'A' : case 'B' : case 'C' : number *= 10; number += 2; break;
case 'D' : case 'E' : case 'F' : number *= 10; number += 3; break;
case 'G' : case 'H' : case 'I' : number *= 10; number += 4; break;
case 'J' : case 'K' : case 'L' : number *= 10; number += 5; break;
case 'M' : case 'N' : case 'O' : number *= 10; number += 6; break;
case 'P' : case 'Q' : case 'R' : case 'S' : number *= 10; number += 7; break;
case 'T' : case 'U' : case 'V' : number *= 10; number += 8; break;
case 'W' : case 'X' : case 'Y' : case 'Z' : number *= 10; number += 9; break;
}
}
else if (Character.isDigit(ch))
{
number *= 10; number += Character.getNumericValue(ch);
}
else
{
System.out.println("Invalid character!");
}
} // End of for loop
// Return actual number only at the end of the function
return number;
}// End of full_number function
}
輸入輸出:
Enter The Phone Number (With Letters):
1800FLOWERS
Output phone number for '1800FLOWERS' is '18003569377'
盡管已經回答了這個問題,但我想指出一些事情。 不要使用 int 或 long 來保存電話號碼! 您將丟失前導零! 此外,您將很容易超出您的整數或遠程范圍。 另外,目前接受的答案有點難以理解。 我只會使用更少且更容易理解的代碼:
public String toNormalPhoneNumber(String phoneNumber) {
String normal = "";
foreach (char c : phoneNumber.toUppercase().toCharArray())
normal += getKeypadNumber(c);
return normal;
}
public char getKeypadNumber(char characterToConvert) {
if (Character.isDigit(characterToConvert))
return characterToConvert;
else {
switch (characterToConvert) {
case 'A' : case 'B' : case 'C' : return '2';
case 'D' : case 'E' : case 'F' : return '3';
case 'G' : case 'H' : case 'I' : return '4';
case 'J' : case 'K' : case 'L' : return '5';
case 'M' : case 'N' : case 'O' : return '6';
case 'P' : case 'Q' : case 'R' : case 'S' : return '7';
case 'T' : case 'U' : case 'V' : retrun '8';
case 'W' : case 'X' : case 'Y' : case 'Z' : return '9';
default return '?';
}
}
}
我認為這更容易理解。
這是 ac# 答案
class Program
{
static void Main(string[] args)
{
Console.WriteLine("Please enter the phone number (With Letters): ");
string initial_phone_number = Console.ReadLine();
initial_phone_number = initial_phone_number.ToUpper();
string phone_number_final = full_number(initial_phone_number);
Console.WriteLine("Output phone number for " + initial_phone_number + " is " + phone_number_final);
Console.ReadLine();
}
public static string full_number(String initial_phone_number)
{
string number = "";
string digit = "";
int strLen = initial_phone_number.Length;
for (int currCharacter = 0; currCharacter < strLen; currCharacter++)
{
string ch = initial_phone_number.Substring(currCharacter,1);
int n;
bool isNumeric = int.TryParse(ch, out n);
if (!isNumeric)
{
switch (ch)
{
case "A": case "B": case "C": digit = "2"; break;
case "D": case "E": case "F": digit = "3"; break;
case "G": case "H": case "I": digit = "4"; break;
case "J": case "K": case "L": digit = "5"; break;
case "M": case "N": case "O": digit = "6"; break;
case "P": case "Q": case "R": case "S": digit = "7"; break;
case "T": case "U": case "V": digit = "8"; break;
case "W": case "X": case "Y": case "Z": digit = "9"; break;
}
number = number + digit;
}
else if (isNumeric)
{
number = number + n.ToString();
}
else
{
Console.WriteLine("Invalid character!");
}
}
return number;
}
}
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