Question

编写一个读取整数的程序，查找其中最大的整数，并计算其出现次数。假设输入以数字0结尾。假设您输入3 5 2 5 5 5 0;程序发现最大值为5,5的出现次数为4。

设计程序，允许用户重新运行   在同一次运行中使用不同输入的程序。

public void findLargestInteger(){


       //create Scanner object
          Scanner input = new Scanner(System.in);
          int x;


        do { 
       //prompt user input
          System.out.print("Enter an integer, the input ends if it is 0:");

       //declare variables
          int n, countNeg = 0, countPos = 0;
          float sum = 0;


       //calculate how many positive and negative values, total, and average
          while ((n = input.nextInt()) != 0) {
             sum = sum + n;

             if (n > 0) {
                countPos++;
             } 
             else if (n < 0) {
                countNeg++;
             }

          }

         //display results 
          if (countPos + countNeg == 0) {
             System.out.println("No numbers are entered except 0");
             System.exit(0);
          }

          System.out.println("The number of positives is " + countPos);
          System.out.println("The number of negatives is " + countNeg);
          System.out.println("The total is " + sum);
          System.out.println("The average is " + (sum / (countPos + countNeg)));

       }while ((x = input.nextInt()) != 0);
}

如何在最后提示正确显示并保留它运行？

输出：

Enter an integer, the input ends if it is 0:
1 2 3 0 
The number of positives is 3 
The number of negatives is 0 
The total is 6.0
The average is 2.0 
1 
Enter an integer, the input ends if it is 0:
1 2 3 0 
The number of positives is 3 
The number of negatives is 0 
The total is 6.0 
The average is 2.0 
1 
Enter an integer, the input ends if it is 0:
2 3 4 0 
The number of positives is 3 
The number of negatives is 0
The total is 9.0 The average is 3.0 
1 
Enter an integer, the input ends if it is 0:
2 3 4 0 
The number of positives is 3 
The number of negatives is 0 
The total is 9.0 
The average is 3.0

Answer 1

如果您真的想继续重复计划，可以将while((x = input.nextInt()) != 0);更改为while(true);。这是一个无限循环，虽然这不是一个好方法。

因此，您可以编写类似

的内容，而不是寻找下一个整数并与0进行比较

System.out.print("Do you want to quit? (y/n): ");

在你的循环结束时（就在while((x = input.nextInt()) != 0)行之前）。

然后不检查0但是y。至少那时你没有让程序等待用户输入内容而不知道发生了什么。

编辑：或者你可以使用一个计数器，如果你想在它终止之前运行它两次或三次;）

Answer 2

您是否尝试过在while循环中调用函数？

public void findLargestInteger() {
    // Your code
}

public static void main(String[] args) {
    do {
        findLargestInteger();
        Scanner reader = new Scanner(System.in);  // Reading from System.in
        System.out.println("Would you like to continue? (0/1) ");
        int n = reader.nextInt();
    } while(n == 1);
}

在同一次运行中重新运行该程序

2 个答案: