Java Codelab逻辑错误

Question

有人可以解释为什么我会遇到逻辑错误。 输出应该是 输入·浇道·1·名称：·输入·浇道·1·时间·（在·分钟）：·输入·浇道·2·名称：·（在·分钟）·输入·浇道·2·时间：·输入·亚军·3·名称：·输入·亚军·3·时间·（在·分）：·Runner3↵ Runner1↵ Runner2↵

import java.util.Scanner;

public class Runners {

public static void main(String[] args) {
    Scanner keyboard = new Scanner(System.in);

    double  t1,
            t2,
            t3;



    System.out.print("Enter runner 1 name:");
    String name1;
    name1 = keyboard.nextLine();
    System.out.print(" Enter runner 1 time (in minutes):");
    String time1;
    time1 = keyboard.nextLine();

    System.out.print(" Enter runner 2 name:");
    String name2;
    name2 = keyboard.nextLine();
    System.out.print(" Enter runner 2 time (in minutes):");
    String time2;
    time2 = keyboard.nextLine();

    System.out.print(" Enter runner 3 name:");
    String name3;
    name3 = keyboard.nextLine();
    System.out.print(" Enter runner 3 time (in minutes):");
    String time3;
    time3 = keyboard.nextLine();


    t1 = Double.parseDouble(time1);
    t2 = Double.parseDouble(time2);
    t3 = Double.parseDouble(time3);




    if(t1 > t2 && t1 > t3){
        if(t2>t3){
            System.out.print(" "+ name3 + "\n"+ name2 + "\n" + name1+ "\n");
        }
        else{
            System.out.print(" "+ name2 + "\n" + name3 + "\n" + name1+ "\n");
        }
    }
    else if(t2>t1 && t2>t3){
        if(t1>t3){
            System.out.print(" " + name3 + "\n" + name1 + "\n" + name2+ "\n");
        }
        else{
            System.out.print(" " + name1 + "\n" + name3 + "\n" + name2+ "\n");
        }
    }
    else if(t3>t2 && t3>t1){
        if(t2>t1){
            System.out.print(" " + name1 + "\n" + name2 + "\n" + name3+ "\n");
        }
        else{
            System.out.print(" " + name2 + "\n" + name1 + "\n" + name3+ "\n");
        }
    }
    else if(t3==t2 && t3==t1){
        if(t2>t1){
            System.out.print(" " + name1 + "\n" + name2 + "\n" + name3+ "\n");
        }
        else{
            System.out.print(" " + name1 + "\n" + name2 + "\n" + name3 + "\n");
        }




}
    }
}

Answer 1

我来到这里寻找同一问题的答案，并在这个例子中探讨了代码。最后有一个逻辑问题。第67行中的AND应为OR。你不能让条件1等于2而2等于3，然后询问2是否大于1并期望它能够工作。

    import java.util.Scanner;

public class Runners {

public static void main(String[] args) {
    Scanner keyboard = new Scanner(System.in);

    double  t1,
            t2,
            t3;



    System.out.print("Enter runner 1 name:");
    String name1;
    name1 = keyboard.nextLine();
    System.out.print(" Enter runner 1 time (in minutes):");
    String time1;
    time1 = keyboard.nextLine();

    System.out.print(" Enter runner 2 name:");
    String name2;
    name2 = keyboard.nextLine();
    System.out.print(" Enter runner 2 time (in minutes):");
    String time2;
    time2 = keyboard.nextLine();

    System.out.print(" Enter runner 3 name:");
    String name3;
    name3 = keyboard.nextLine();
    System.out.print(" Enter runner 3 time (in minutes):");
    String time3;
    time3 = keyboard.nextLine();


    t1 = Double.parseDouble(time1);
    t2 = Double.parseDouble(time2);
    t3 = Double.parseDouble(time3);




    if(t1 > t2 && t1 > t3){
        if(t2>t3){
            System.out.print(" "+ name3 + "\n"+ name2 + "\n" + name1+ "\n");
        }
        else{
            System.out.print(" "+ name2 + "\n" + name3 + "\n" + name1+ "\n");
        }
    }
    else if(t2>t1 && t2>t3){
        if(t1>t3){
            System.out.print(" " + name3 + "\n" + name1 + "\n" + name2+ "\n");
        }
        else{
            System.out.print(" " + name1 + "\n" + name3 + "\n" + name2+ "\n");
        }
    }
    else if(t3>t2 && t3>t1){
        if(t2>t1){
            System.out.print(" " + name1 + "\n" + name2 + "\n" + name3+ "\n");
        }
        else{
            System.out.print(" " + name2 + "\n" + name1 + "\n" + name3+ "\n");
        }
    }
    else if(t3==t2 || t3==t1){
        if(t2>t1){
            System.out.print(" " + name1 + "\n" + name2 + "\n" + name3+ "\n");
        }
        else{
            System.out.print(" " + name1 + "\n" + name2 + "\n" + name3 + "\n");
        }




}
    }
}