reading an array of character (String) in java

Question

I am trying to solve a problem with Java and the input is a string of 15 characters for example. when I use a scanner to read it using the method nextLine() it doesn't work because the input string doesn't have a \n in the end for example the input is :

15 3
cccaabababaccbc

my code :

Scanner s = new Scanner(System.in);
int n = s.nextInt();
int k = s.nextInt();
String temp = s.nextLine();
temp = s.next();

note that I read the temp twice to get the \n in the end of the first line.the problem link and my submission is here

I don't understand your problem. Do you have issues reading from user input or for traversing the characters in the String? — Luiggi Mendoza, Jun 21 '18 at 22:34
i mean it just produces a runtime error because it waits too much to get to the end of the string but simply it doesn't end with \n — A.khaled, Jun 21 '18 at 22:36
If you want to read 15 `char`s, read `char`s and not lines (which are `String`s). — Turing85, Jun 21 '18 at 22:40
E.g. [`InputStreamReader#read(char[], int, int)`](https://docs.oracle.com/javase/10/docs/api/java/io/InputStreamReader.html#read(char%5B%5D,int,int)). — Turing85, Jun 21 '18 at 22:43
Possible duplicate of [Scanner is skipping nextLine() after using next() or nextFoo()?](https://stackoverflow.com/questions/13102045/scanner-is-skipping-nextline-after-using-next-or-nextfoo) — Johannes Kuhn, Jun 21 '18 at 23:06

score 0 · Answer 1 · answered Jun 22 '18 at 00:26

You can read two numbers with nextInt() and then read string with only next() - it should work fine. Scanner next() will ignore whitespaces - see the init string in my example.

public static void main(String [] args){
   Scanner scanner = new Scanner("3 5 \n abcdef    z");
   System.out.println(scanner.nextInt());
   System.out.println(scanner.nextInt());
   System.out.println(scanner.next());
   System.out.println(scanner.next());
}

it gives me result

3
5
abcdef
z

reading an array of character (String) in java

1 Answers1