[2019.01.03] Exceptions - String to Integer

Problem

Objective 
Today, we're getting started with Exceptions by learning how to parse an integer from a string and print a custom error message. Check out the Tutorial tab for learning materials and an instructional video!

Task 
Read a string, , and print its integer value; if  cannot be converted to an integer, print Bad String.

Note: You must use the String-to-Integer and exception handling constructs built into your submission language. If you attempt to use loops/conditional statements, you will get a  score.

Input Format

A single string, .

Constraints

• , where  is the length of string .
•  is composed of either lowercase letters () or decimal digits ().

Output Format

Print the parsed integer value of , or Bad String if  cannot be converted to an integer.

Sample Input 0

3

Sample Output 0

3

Sample Input 1

za

Sample Output 1

Bad String

Explanation

Sample Case  contains an integer, so it should not raise an exception when we attempt to convert it to an integer. Thus, we print the . 
Sample Case  does not contain any integers, so an attempt to convert it to an integer will raise an exception. Thus, our exception handler prints Bad String.

How I solved the problem

# String으로 입력받은 문자를 int로 변환하면서 parse Exception이 나타나도록 구성해 주는 것이 핵심.

# excetion이 일어났을 때를 보려면 try ~ catch 구문을 사용!

import java.io.*;
import java.util.*;
import java.text.*;
import java.math.*;
import java.util.regex.*;
 
public class Solution {
 
    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        String S = in.next();
 
        try {
            int result = Integer.parseInt(S); // 입력받은 String을 Int로 변환
            System.out.println(result);
        } catch (Exception e) {
            System.out.println("Bad String");
        }
    }
}

[출처 : https://www.hackerrank.com ]