java如何一次将用户输入与多个变量进行比较？

Question

我一直在尝试一个基本的程序，它复制了游戏《智囊团》。到目前为止，我已经能够得到4个数据点，4个用户输入作为一个整数，但是我在比较4个数据点和4个用户输入时遇到了一个问题。目标是有4个RNG（范围从1到9），用户一次猜4个数字。他得到的唯一提示是一个数字是正确的。到目前为止，我已经能够生成一个低效的代码，设置为比较4个变量，例如（a=rng1，b=rng2，c=rng3，d=rng4）和用户输入。如何有效地将用户输入与生成的所有RNG进行比较。在这个例子中，我希望能够比较s1和poop；poop2；poop3；还有便便。我不想再为此编写100行代码。提前谢谢。代码如下

import java.util.*;
public class RandomNumberGenerator{;
private static Scanner in;
public static void main(String[] args){
  Random r = new Random();
  float s;
  in = new Scanner(System.in);
  int poop = (r.nextInt(10-1)+1);
  int poop1 = (r.nextInt(10-1)+1);
  int poop2 = (r.nextInt(10-1)+1);
  int poop3 = (r.nextInt(10-1)+1);
for (int i = 0; i < 80; i++)
{
System.out.println("Enter a number from 1-9");
s = in.nextFloat();
float s1 = in.nextFloat();
float s2 = in.nextFloat();
float s3 = in.nextFloat();

if(poop == s && poop1 == s1 && poop2 == s2 && poop3 == s3) {
    System.out.println("You are correct HeHe XD");
    break;
    }
if(poop > s && poop1 == s1 && poop2 == s2 && poop3 == s3){
    System.out.println("You have 1 wrong");
}
if(poop < s && poop1 == s1 && poop2 == s2 && poop3 == s3){
    System.out.println("You have 1 wrong");
}
if(poop > s && poop1 == s1 && poop2 == s2 && poop3 == s3){
    System.out.println("You have 1 wrong");
}
if(poop == s && poop1 > s1 && poop2 == s2 && poop3 == s3){
    System.out.println("You have 1 wrong");
}
if(poop == s && poop1 < s1 && poop2 == s2 && poop3 == s3){
    System.out.println("You have 1 wrong");
}
if(poop == s && poop1 == s1 && poop2 > s2 && poop3 == s3){
    System.out.println("You have 1 wrong");
}
if(poop == s && poop1 == s1 && poop2 < s2 && poop3 == s3){
    System.out.println("You have 1 wrong");
}
if(poop == s && poop1 == s1 && poop2 == s2 && poop3 > s3){
    System.out.println("You have 1 wrong");
}
if(poop == s && poop1 == s1 && poop2 == s2 && poop3 < s3){
    System.out.println("You have 1 wrong");
}
if(poop != s && poop1 != s1 && poop2 == s2 && poop3 == s3){
    System.out.println("You have 2 wrong");
}
if(poop != s && poop1 == s1 && poop2 != s2 && poop3 == s3){
    System.out.println("You have 2 wrong");
}
if(poop != s && poop1 == s1 && poop2 == s2 && poop3 != s3){
    System.out.println("You have 2 wrong");
}
if(poop == s && poop1 != s1 && poop2 != s2 && poop3 == s3){
    System.out.println("You have 2 wrong");
}
if(poop == s && poop1 != s1 && poop2 == s2 && poop3 != s3){
    System.out.println("You have 2 wrong");
}
if(poop == s && poop1 != s1 && poop2 != s2 && poop3 != s3){
    System.out.println("You have 3 wrong");
}
if(poop != s && poop1 == s1 && poop2 != s2 && poop3 != s3){
    System.out.println("You have 3 wrong");
}
if(poop != s && poop1 != s1 && poop2 == s2 && poop3 != s3){
    System.out.println("You have 3 wrong");
}
if(poop != s && poop1 != s1 && poop2 != s2 && poop3 == s3){
    System.out.println("You have 3 wrong");
}
if(poop != s && poop1 == s1 && poop2 != s2 && poop3 != s3){
    System.out.println("You have 3 wrong");
}
if(poop != s && poop1 != s1 && poop2 != s2 && poop3 != s3){
    System.out.println("You have 4 wrong");
}

 }

 }
}

Answer 1

您可以使用两个ArrayList来存储4个数据点和用户输入。然后使用ArrayList.contains()进行比较。你不必写太多if condition。请参阅下面我的代码：

public static void main(String[] args) {
      final int totalPoints = 4;
      Random r = new Random();
      in = new Scanner(System.in);

      ArrayList<Integer> poops = new ArrayList<Integer>(totalPoints);
      ArrayList<Integer> userinputs = new ArrayList<Integer>(totalPoints);

      for (int i=0; i<totalPoints; i++) {
          int poop = (r.nextInt(10-1)+1);
          System.out.println("Random datapoints: " + poop);
          poops.add(poop);
      }


      System.out.println("Enter a number from 1-9");
      for (int i=0; i<totalPoints; i++) {
          int s = in.nextInt();
          userinputs.add(s);
      }

      int correct = 0;
      for (int i=0; i<4; i++) {
          if (poops.contains(userinputs.get(i))) {
              correct++;
          }
      }

      System.out.println("You have " + correct + "correct numbers");

}

Answer 2

将变量放在List中，然后可以使用contains检查元素是否在List中。 实际上，你会得到两个数字作为智囊团的反馈，有多少是正确的数字，有多少也在正确的位置。 由于mastermind中可能存在重复项，您可以对原始List进行copy，并从List中为每个正确的guess{}创建一个copy。否则，如果主控contains这个数字，猜测所有相同的number是正确的

int size = 4;
List<Integer> poops = new ArrayList<>();
for(int i =0; i<size; i++){
    poops.add((r.nextInt(10-1)+1));
}
...
List<Integer> guesses= new ArrayList<>();
for(int i =0; i<size; i++){
   guesses.add(in.nextInt());
}
int correctNumberCount = 0;
List<Integer> poopsCopy = new ArrayList<>(poops);
for(Integer guess: guesses){
    if(poopsCopy.contains(guess)){
        correctNumberCount ++;
        //For handling duplicates
        poopsCopy.remove(guess);
    }   
} 
int correctCount = 0;
int counter = 0;
for(Integer guess: guesses){
    if(guess == poops.get(counter)){
        correctCount++;
    }
    counter++; 
}
//Inform the user about correctNumberCount and correctCount 

Answer 3

根据您的要求，您只需要获得每个输入组合的错误编号，而不需要输入错误的位置。所以，你没有必要列举所有的可能性

您可以尝试下面的代码，size是一个数字，指示用户可以输入多少

我用List来存储数字。把一系列数据保存在容器中是一个好习惯，这样你就可以用loop轻松地操作它们，loop是计算机最强大的功能之一

import java.util.*;

public class RandomNumberGenerator {

    private static Scanner in;

    private static List<Integer> expected = new ArrayList<Integer>();
    private static List<Integer> input = new ArrayList<Integer>();

    static private final int size = 4;

    static int correctNumber = 0;

    public static void check(List<Integer> expected, List<Integer> input) {
        for (int i = 0; i < size; i++) {
            if (expected.get(i) == input.get(i)) {
                correctNumber++;
            }
        }

        if (correctNumber == size) {
            System.out.println("You are correct HeHe XD");
            return;
        } else {
            int wrongNumber = size - correctNumber;
            System.out.println("You have " + wrongNumber + " wrong");
            correctNumber = 0;
        }
    }

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

        int count = size;
        while(count > 0){
            expected.add((r.nextInt(10 - 1) + 1));
            count--;
        }

        for (int i = 0; i < 80; i++) {
            System.out.println("Enter 4 numbers from 1-9");
            count = size;
            while(count > 0){
                input.add(in.nextInt());
                count--;
            }
            check(expected, input);
        }
    }
}

Answer 4

那么：

import java.util.*;
public class RandomNumberGenerator{;
private static Scanner in;
public static void main(String[] args){
  Random r = new Random();
  float s;
  in = new Scanner(System.in);
  int[] poops = new int[4];
  for(int i=0;i<poops.length;i++){
      poops[i]=r.nextInt(10-1)+1;
  }

float[] s = new float[4];

for (int i = 0; i < 80; i++){
    System.out.println("Enter a number from 1-9");
    int error =0;
    for(int j=0;j<poops.length;j++){
         if(poop[j] != in.nextFloat()){
             error++;
         }
     }
     if(result == 0){
         System.out.println("You are correct HeHe XD");
         break;
     }else{
         System.out.println("You have "+error+" wrong");
     }
 }

}

Answer 5

你不能用名为p1，p2，p3

请后退一步，阅读有关数组的概念

然后创建两个数组（最好使用int；而不是float！）。 第一个数组携带程序想出的4个值；第二个数组由来自用户的4个值填充

然后你迭代两个数组；来计算“用户”数组中有多少元素在“计算机”数组中给出。真的：你永远都不应该写包含那么多if/else语句的代码。这样的代码绝对不可能读取或维护。不要那样做。永远

或者，您可以查看集合，它们是更复杂的数据结构。在本例中，可以使用整数对象的列表/数组列表。当你对这些东西感兴趣时；使用你最喜欢的搜索引擎并阅读相关内容