Question

确定。所以我正在尝试运行一个程序，但它并没有那么好用。我想我很接近，却错过了一步。

我要求用户输入一个短语（完全随机），然后我要求他们输入另一个短语，即三个单词，所有单词都在一起。该程序的目标是打印出第一个短语，但在第一个位置的第二个短语中具有第三个单词的位置。我还希望得到第一和第三字。

User input : My name is George
user input: name is George

desired output: my GEORGE is NAME

这就是我所拥有的。

import java.util.Scanner;
public class Assign3 {

    public static void main(String[] args) {    
        String one, two, three,four, five,six,seven,eight;
        Scanner kbrd = new Scanner(System.in);
        System.out.println("please enter a phrase, and then pick three words that you want to switch");
        one= kbrd.nextLine();

        two= kbrd.next();
        three = kbrd.next();
        four = kbrd.next();
        seven = two+three+four;

        System.out.println(one.indexOf(two));
        System.out.println(one.lastIndexOf(four));

        five = four.toUpperCase()+three+two.toUpperCase();
        System.out.println(five);

        eight = one.replace(seven,five);
        System.out.println(eight);
    }
}

Answer 1

你几乎得到了你所缺少的几个空格，因为它没有找到你的字符串并替换它。

新代码应如下所示：

public static void main(String[] args) {    
String one, two, three,four, five,six,seven,eight;
      Scanner kbrd = new Scanner(System.in);
      System.out.println("please enter a phrase, and then pick three words that you want to switch");
      one= kbrd.nextLine();

      two= kbrd.next()+" "; // Added space
      three = kbrd.next()+" "; // Added space
      four = kbrd.next()+" "; // Added space
      seven = two+three+four;

      System.out.println(one.indexOf(two));
      System.out.println(one.lastIndexOf(four));

      five = four.toUpperCase()+three+two.toUpperCase();
      System.out.println(five);

      eight = one.replace(seven.trim(),five.trim()); // timing the last space
      System.out.println(eight);
}

输出：

please enter a phrase, and then pick three words that you want to switch
My name is George
name is George
3
-1
GEORGE is NAME 
My GEORGE is NAME

将用户输入作为字符串并重新排列单词的顺序？

1 个答案: