I use it all the time on hackerrank/leetcode BufferedReader br = new BufferedReader(new InputStreamReader(System.in)); String lines = br.readLine(); String[] strs = lines.trim().split(“\\s+”); for (int i = 0; i < strs.length; i++) { a[i] = Integer.parseInt(strs[i]); }
As per the javadoc for Scanner: When a scanner throws an InputMismatchException, the scanner will not pass the token that caused the exception, so that it may be retrieved or skipped via some other method. That means that if the next token is not an int, it throws the InputMismatchException, but the token stays there. … Read more
Use an URL instead of File for any access that is not on your local computer. URL url = new URL(“http://www.puzzlers.org/pub/wordlists/pocket.txt”); Scanner s = new Scanner(url.openStream()); Actually, URL is even more generally useful, also for local access (use a file: URL), jar files, and about everything that one can retrieve somehow. The way above interprets … Read more
The simplest thing is to not close Scanner if you don’t want to close the underlying stream. Ideally you should create just one Scanner which you use for the life of the program. In any case, it appears you don’t have a good reason to close it.
Please stop writing faulty CSV parsers! I’ve seen hundreds of CSV parsers and so called tutorials for them online. Nearly every one of them gets it wrong! This wouldn’t be such a bad thing as it doesn’t affect me but people who try to write CSV readers and get it wrong tend to write CSV … Read more
Overview of Scanner.hasNextXXX methods java.util.Scanner has many hasNextXXX methods that can be used to validate input. Here’s a brief overview of all of them: hasNext() – does it have any token at all? hasNextLine() – does it have another line of input? For Java primitives hasNextInt() – does it have a token that can be … Read more
with Scanner you need to check if there is a next line with hasNextLine() so the loop becomes while(sc.hasNextLine()){ str=sc.nextLine(); //… } it’s readers that return null on EOF ofcourse in this piece of code this is dependent on whether the input is properly formatted
That’s because the Scanner.nextInt method does not read the newline character in your input created by hitting “Enter,” and so the call to Scanner.nextLine returns after reading that newline. You will encounter the similar behaviour when you use Scanner.nextLine after Scanner.next() or any Scanner.nextFoo method (except nextLine itself). Workaround: Either put a Scanner.nextLine call after … Read more
You have a single extra newline at the end of your file. hasNextLine() checks to see if there is another linePattern in the buffer. hasNext() checks to see if there is a parseable token in the buffer, as separated by the scanner’s delimiter. Since the scanner’s delimiter is whitespace, and the linePattern is also white … Read more
You have to put file extension here File file = new File(“10_Random.txt”);