java - 如何打印一行中第 N 个单词的第 N 个字符？

Question

我正在处理一个家庭作业问题，该问题要求在同一行上打印第 N 个字符和第 N 个单词且没有空格。如果第 N 个单词太短且没有第 N 个字符，程序将打印该单词的最后一个字符。如果用户输入一个空词(简单的按下)，该词将被忽略。

(我们还没有学习方法，所以我不应该使用它们)

请参阅下面的代码，如果它没有第 N 个字符，我不确定如何让我的代码打印该单词的最后一个字符。

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();
    }

}
}

当我运行它时出现错误:

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)

我希望在同一行看到第 N 个单词和第 N 个字符

Answer 1

用户输入也可以包含少于 N 个单词，对吗？首先检查应该是这样。

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();
}