在 Java 中只打印一次随机数组变量

Question

我有一个 uni 的任务，我必须创建一个 Java 程序来生成随机配对的客户及其交易总额。

它分为两部分，第一部分产生客户及其交易，如下所示：

Customer ID, Transaction Value
1, 74.36 
1, 44.98 
3, 6.44
0, 19.13
3, 59.44
2, 81.56
0, 87.21
4, 40.9
1, 42.11
3, 66.05

第二个总结了每个客户的交易总数，如下所示：

Customer: 1 Transactions: 3.0
Customer: 1 Transactions: 3.0
Customer: 3 Transactions: 3.0
Customer: 0 Transactions: 2.0
Customer: 3 Transactions: 3.0
Customer: 2 Transactions: 1.0
Customer: 0 Transactions: 2.0
Customer: 4 Transactions: 1.0
Customer: 1 Transactions: 3.0
Customer: 3 Transactions: 3.0

我的问题是第二部分应该只生成每个客户 ID 一次，即1,3,0,2,4。我只能使用 int 和双精度变量来做到这一点，而不创建任何其他数组或结构。我的代码可以在下面找到。

 import java.util.*;
 public class Assignment3 {
   public static long studentNumber=1234567; 
   public static int customerID[];
   public static double transactionValue[];
   public static void initialiseData(int size) {
     customerID = new int[size];
     transactionValue = new double[size];
     Random rand = new Random(studentNumber);
     for (int i=0; i<size; i++) {
       customerID[i] = rand.nextInt(size / 2);
       transactionValue[i] = rand.nextInt(10000) / 100.0;
     }
   }
   public static void main(String args[]) {
     int size=10;
    initialiseData(size);
    // Your code should only be below this line
    double transaction = 0;
    int customer = 0;
    int customer_Total = 0;
    int count = 0;
    int customer_count = 0;
    double transaction_Total = 0;
    System.out.println("Customer ID, Transaction Value");
    for (size= 0; size < customerID.length; size++) {
    customer= customerID[size];
    transaction= transactionValue[size];
    System.out.println(customer + ", " + transaction);
    }
    System.out.println("");
    for(customer_count = 0; customer_count < customerID.length; customer_count++) {
      transaction_Total= 0;
      customer_Total = customerID[customer_count];
      count = customerID[customer_count];
      //System.out.println(count);
      for (int customer_count2 = 0;
           customer_count2 < customerID.length;
           customer_count2++) {                         
        if (customer_Total == customerID[customer_count2]) {
          transaction_Total++;
          //System.out.println(customer_count2);
        }
      }
      System.out.println("Customer: "+ customer_Total + " " +
                         "Transactions: " + transaction_Total); 
    }
    // Your code should not be below this line
  }
}

Answer 1

我建议你看看Collections.shuffle是如何工作的。 在您的情况下，您可以创建一个包含所有可能值的数组，然后以随机顺序"洗牌"。

Answer 2

编辑：在此代码中：

for(customer_count = 0; customer_count < customerID.length; customer_count++) {
  transaction_Total= 0;
  customer_Total = customerID[customer_count];
  count = customerID[customer_count];
  //System.out.println(count);
  for (int customer_count2 = 0;
       customer_count2 < customerID.length;
       customer_count2++) {                         
    if (customer_Total == customerID[customer_count2]) {
      transaction_Total++;
      //System.out.println(customer_count2);
    }
  }
  System.out.println("Customer: "+ customer_Total + " " +
                     "Transactions: " + transaction_Total); 
}

在"transaction_Total= 0;"之后添加以下内容：

int wasBefore = 0; //are you allowed to use boolean variables?
for (int customer_count3 = 0;
       customer_count3 < customer_count;
       customer_count3++) {                         
    if (customer_Total == customerID[customer_count3]) {
      wasBefore = 1;
    }
}
if (wasBefore==1) {continue;}

通过这种方式，您仍然可以获得客户的随机顺序，而不会责备他们

Answer 3

所以你的家庭作业有点烦人，因为像不使用数组这样的限制是不寻常的。

但您可以执行以下操作：您的客户 ID 从 0 变为 size/2，因此可以编写一个循环来计算每个可能客户的事务。喜欢这个

for(int customer_id = 0; customer_id <= size /2 ; customer_id++){
    int transaction_sum = 0;
    for (j= 0; j < customerID.length; j++)
        if(customerID[j] == i)
             transaction_sum++;
    if( transaction_sum > 0)
        System.out.println("Customer: " + customer_id +
                           "  Transactions: " + transaction_sum);
}

但这不是一个很好的解决方案，因为对于大量可能的 ID 来说，这很慢。