How to force Scanner to read unfinished line

Question

I have a problem with last line of input. I cannot force Scanner to read a number from last line since it is unfinished. When I copy input with one more line after the previous last line, there is no problem, but i cannot have the extra line there.

Link to problem

Example of input:

I've tried sc.nextInt(), sc.next(), sc.nextLine() and nothing can reach the -1. Even sc.hasNext, sc.hasNextInt etc. are waiting.

while(true)
    {
        line = sc.nextLine();
        try {
            a = Integer.parseInt(line);
        } 
        catch (NumberFormatException e) {
        }

        if(a == -1)
        {
            break;
        }

        rr = new int[a][2];

        for(int i = 0; i < a; i++)
        {
            line = sc.nextLine();
            numbers = line.split(" ");

            try {
                rr[i][0] = Integer.parseInt(numbers[0]);
                rr[i][1] = Integer.parseInt(numbers[1]);
            } 
            catch (NumberFormatException e) {
            }

        }

        b = sc.nextInt();

        sc.nextLine();

        tt = new int[b][2];

        for(int i = 0; i < b; i++)
        {
            line = sc.nextLine();
            numbers = line.split(" ");

            try {
                tt[i][0] = Integer.parseInt(numbers[0]);
                tt[i][1] = Integer.parseInt(numbers[1]);
            } 
            catch (NumberFormatException e) {
            }

        }
        sc.nextLine();

    }

Read all lines with nextLine. Parse the Integer(s) as a secondary step — OneCricketeer, Nov 27 '16 at 20:54
tried, unfortunately it didnt work... probably it waits for line ending or something — Daniel Švarc, Nov 27 '16 at 20:54
try this http://stackoverflow.com/questions/5868369/how-to-read-a-large-text-file-line-by-line-using-java should help — Maytham Fahmi, Nov 27 '16 at 20:55
Maybe try `sc.setDelimiter("\\s|$")` so it'll catch the end of the string. — shmosel, Nov 27 '16 at 21:08
Didn't help... It works on my pc now but UVA Online Judge will not take it — Daniel Švarc, Nov 27 '16 at 21:11
You want a `continue` within the catch statements. Also, not really clear what the `b` value is supposed to be. — OneCricketeer, Nov 27 '16 at 21:12
it does not really matter since any exception means fail... b means number of something, I need it to make a field of preffered size — Daniel Švarc, Nov 27 '16 at 21:13
When you read the blank line then call `parseInt(line)`, it'll throw `NumberFormatException`, but since you catch **and ignore** the exception, what value do you think `a` has? *(Hint: It's unchanged)*. — Andreas, Nov 27 '16 at 21:16
Problem itself is easy, I have problem with validation :D https://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=3811 — Daniel Švarc, Nov 27 '16 at 21:22

denvercoder9 · Answer 1 · 2016-11-28T06:16:09.127

1

while(true) {
    // read the line.. 
    line = sc.nextLine().trim();
    // check if it's a blank line...
    // blank line means new test case.. Take the value of n
    if (line.length() == 0) line = sc.nextLine().trim();

    // now parse the value of n.. If it's -1 then end of input
    a = Integer.parseInt(line);
    if (a==-1) break;

    // rest of your input
}

Take the value of b like this:

b = Integer.parseInt(sc.nextLine().trim());

And remove the 'extra' sc.nextLine()from below of b and from the bottom of while loop. That case is handled at the top.

edited Nov 28 '16 at 06:16

answered Nov 27 '16 at 21:32

denvercoder9

2,979
3
28
41

Doesnt work neither. I tried to debug it and it stucks on the if(line.length() == 0) line = sc.nextLine(); – Daniel Švarc Nov 27 '16 at 21:55
Use these sites: http://uhunt.felix-halim.net/id/377362 https://www.udebug.com/UVa/12389 I have edited my answer. You need to probably trim the strings before you parse it. – denvercoder9 Nov 28 '16 at 06:15

score 0 · Answer 2 · answered Nov 27 '16 at 21:19

0

Before trying to get the -1 you could have a separate scanner use the delimiter /n (line break) and have it delete the extra break between -1 and the last line before it. Then you could go through and just scan with .nextLine() or .nextInt().

answered Nov 27 '16 at 21:19

Dane Williams

82
12

Isnt it the same what does .nextLine() on the same place? – Daniel Švarc Nov 27 '16 at 21:21
.nextLine() will return whatever is on the next line. What I am proposing is to set up a scanner so that it searches for blank lines, and then edits the file so that it contains no blank lines. After that, it goes through and gets the integers you want, and should have no problem getting -1 since it will no longer have an empty line separating it from the other numbers. – Dane Williams Nov 27 '16 at 21:27

How to force Scanner to read unfinished line

2 Answers2