基本的Java:猜数字二进制搜索 [英] Basic java: Guess the number binary search

问题描述

好吧，因此用户认为(例如，不键入)介于1到10之间的数字，程序会提示用户您的数字是否小于或等于X?"，然后用户键入对或错.到目前为止，我已经设法进行了搜索间隔，但是我不知道如何继续.主要问题是，如果允许我使用正确！"，则只能使用true或false.那就没问题了.

Alright, so the user thinks (does not type it) of a number between 1 and 10 (for example), the program prompts the user "Is you number less than or equal to X?", then the user types either true or false. So far, I have managed to do the search interval, but I have no idea how to continue. Main problem is, I am only allowed to use true of false, if I was allowed to use "correct!" then there will be no problem.

import java.util.*;

public class GuessTheNumber {

public static void main (String[]args){

    Scanner scan = new Scanner (System.in);
    System.out.println("Think of a number between 1 and 10\n");

    double max = 10;
    double x = max/2;

    while (true){
        System.out.println("Is your number less than or equal to "+(int)x+" ?");
        String truth = scan.next();

        if(truth.equals("true")){
            x=x/2;
        }
        if(truth.equals("false")){
            x+=x/2;
        }
    }
    //The program breaks at some point
    System.out.println("Your number is: ");
 }
}

程序希望用户键入true或false，因此我们可以忽略其他任何内容.

The program expects the user to type either true or false, so we can ignore anything else.

推荐答案

在 while 循环的开始(内部)添加以下内容:

Add the following at the beginning (inside) of your while loop:

System.out.println("Is your number " + x +" ?");
String truth = scan.next();
if(truth.equals("true")){
    break;
}

其他选项:将您的 if 语句转换为以下形式:

Other option: transform your if statements to look like the following:

if(truth.equals("true")){
    if(x == x/2) break;
    x=x/2;
}
if(truth.equals("false")){
    if(x == x + x/2) break;
    x+=x/2;
}

，然后将最后一条输出行更改为:

and then change the last output line to:

 System.out.println("Your number is: " + x);

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