如何打印一行中第 N 个单词的第 N 个字符？

Question

I am working on a homework problem which requests that print the Nth character the Nth word on the same line with no spaces. If the Nth word is too short and does not have Nth character, the program shall print the last character of that word. If the user enters an empty word (simple presses), that words shall be disregarded.

(We haven't learned methods yet so I am not supposed to use them)

See the code below, I am not sure how to get my code to print the last character of that word if it does not have the Nth character.

import java.util.Scanner;

public class Words {

    public static void main(String[] args) {
        final int N=5;
        Scanner input = new Scanner(System.in);
        System.out.print("Enter a line of words seperated by spaces ");
        String userInput = input.nextLine();
        String[] words = userInput.split(" ");
        String nthWord = words[N];

        for(int i = 0; i < nthWord.length();i++) {
            if(nthWord.length()>=N) {
                char nthChar = nthWord.charAt(N);
                System.out.print("The " + N + "th word in the line entered is " + nthWord + "The " + N + "th charecter in the word is " + nthChar);
            }
            if(nthWord.length()<N) {
                    char nthChar2 = nthWord.charAt(nthWord.length()-1);
                    System.out.print("The " + N + "th word in the line entered is " + nthWord + "The " + N + "th charecter in the word is " + nthChar2);
        }
        input.close();
    }

}
}

When I run this I get an error:

Exception in thread "main" java.lang.StringIndexOutOfBoundsException: String index out of range: 5
    at java.base/java.lang.StringLatin1.charAt(StringLatin1.java:47)
    at java.base/java.lang.String.charAt(String.java:702)
    at Words.main(Words.java:24)

I expect to see the Nth word and the Nth character on the same line

Answer 1

User input can also contain less than N words, right? First check should be that.

public static void main(String[] args) {
    Scanner input = new Scanner(System.in);
    System.out.print("Enter a line of words seperated by spaces ");
    String userInput = input.nextLine();
    String[] words = userInput.split(" ");
    int n = words.length();
    System.out.print("Enter lookup word - N");
    int askedFor = input.nextInt();
    if (askedFor > n) {
        //your logic for this condition
        return;
    }
    String nthWord = words[askedFor-1];
    if (nthWord.length() < askedFor) print(nthWord.charAt(nthWord.length()-1));
    else print(nthWord.charAt(askedFor-1));
    input.close();
}

Answer 2

//Considering line as your input
String[] words = line.split(" ");

//Check if Nth word exists
if(words.length < N){
 System.out.println("Nth word does not exists");
 return;
}

//Check if Nth character exists in Nth word
if(words[N-1].length() < N){
 System.out.println("Nth character in Nth word does not exists");
 return;
}

// returning Nth character from Nth word
// Here N-1 = N; as programming starts with 0th index
return words[N-1].charAt(N-1);

Answer 3

Streams in Java could be a bit harder but, why not?!

You can define a generic function working for any sequence:

static <T> T nthOrLastOrDefault(Collection<T> xs, int nth, T defaultValue) {
    return xs.stream()                                      // sequence as stream
            .skip(nth - 1)                                  // skip n - 1
            .findFirst()                                    // get next (the nth)
            .orElse(xs.stream()                             // or if not available
                    .reduce(defaultValue, (a, b) -> b));    // replace defaultValue with the next and so on
}

Returning the nth element, if not the last one, if not the default value.

Now you simply apply that function for all words and then for the returned word.

(Unfortunately Java manages characters differently and must be converted from integers to characters)

String nthWord = nthOrLastOrDefault(asList(words), N, "");

List<Character> chars = nthWord.chars().mapToObj(i -> (char) i).collect(toList()); // ugly conversion

char nthNth = nthOrLastOrDefault(chars, N, '?');

The output could be

Enter a line of words seperated by spaces:
one two three four five six
The nth char or last or default of the nth word or last or default is 'e'
Enter a line of words seperated by spaces:
one two three four
The nth char or last or default of the nth word or last or default is 'r'
Enter a line of words seperated by spaces:

The nth char or last or default of the nth word or last or default is '?'
Enter a line of words seperated by spaces:

(Complete example code)

static <T> T nthOrLastOrDefault(Collection<T> xs, int nth, T defaultValue) {
    return xs.stream()                                      // sequence as stream
            .skip(nth - 1)                                  // skip n - 1
            .findFirst()                                    // get next (the nth)
            .orElse(xs.stream()                             // or if not available
                    .reduce(defaultValue, (a, b) -> b));    // replace defaultValue with the next and so on
}

public static void main(String[] args) {
    final int N = 5;
    try (final Scanner input = new Scanner(System.in)) {
        while (true) {
            System.out.println("Enter a line of words seperated by spaces:");

            final String userInput = input.nextLine();

            final String[] words = userInput.split(" ");

            String nthWord = nthOrLastOrDefault(asList(words), N, "");

            List<Character> chars = nthWord.chars().mapToObj(i -> (char) i).collect(toList()); // ugly conversion

            char nthNth = nthOrLastOrDefault(chars, N, '?');

            System.out.printf("The nth char or last or default of the nth word or last or default is '%c'%n", nthNth);
        }
    }
}