java - 在另一个 for 循环中使用数组索引值

Question

我正在尝试计算数组中存在唯一数字的次数，使用的索引数量取决于输入的元素数量。除了 1 之外，它主要是可操作的。第一个值没有被考虑在内。循环正在检查 arrays.length -1 ，因此即使我输入，数字 3 也显示为 1 的计数3 两次。我知道解决此问题的最佳方法是运行一个不使用 arrays.length -1 的循环，但是这样我就无法将一个条目与其旁边的条目进行比较，例如 if(a[i] == a[i + 1] && a[i] != 0) 查看某个值是否多次出现。我认为最好的办法是将计数值及其相应的已计数数组值存储在我的 count 方法中，然后在该方法之外执行 for 循环，这可能吗？我看不到一种方法，因为我对 java 相当陌生。我可以有一些指导吗:)

import java.util.Scanner;


public class Prac_ExeOne 
{
        static int count  = 1;
        static int numberUsed = 0; // static its used to show its a class wide variable and there is only one copy.
        static int[] Array =  new int [50]; // the maximum elements inside the array that can be used is 10;
        int size;

              public int fillArray(int[] a)
         {
                 System.out.println("Enter up to " + a.length + " nonnegative numbers.");
                 System.out.println("Mark the end of the list with a negative number.");
                 Scanner keyboard = new Scanner(System.in);

                 int next;
                 int index = 0;
                 next = keyboard.nextInt();
                 while ((next >= 0) && (index < a.length ))  
                 {
                     numberUsed++;
                     a[index] = next;
                     index++;
                     // Print out each value of next
                     System.out.println(next);
                     next = keyboard.nextInt();
                     //System.out.println("Number of indexes used" + numberUsed);
                 }
                 keyboard.close(); // close the keyboard so it can't continue to be used.
                 return index;
        }


            public int[] sort(int[] arrays)
            {

                for(int i = 0;i < arrays.length -1 ;i++ )
                {

                    int store = 0;
                    // Move Larger Values to the right.
                    if (arrays[i + 1 ] < arrays[i])
                    {
                        store = arrays[i];
                        arrays[i] = arrays[i + 1];
                        arrays[i + 1] = store;
                    }
                    // Sort swapped smaller values to the left.
                        for(int j = i; j > 1; j--)
                        {
                           if (arrays[j] < arrays[j - 1])
                           {
                           store = arrays[j];
                           arrays[j] = arrays[j - 1];
                           arrays[j - 1] = store;
                           }
                      }
                }
                return arrays;

            }
            public void count(int[] a)
            {   

                //for each element in array go through if conditons.
                System.out.println("N " + "Count");
                for(int i = 0;i < a.length -1;i++)
                {   
                    if(a[i] == a[i + 1] && a[i] != 0)
                    {
                        count++;

                    }
                    if(a[i] != a[i+1])
                    {
                        count = 1;
                    }
                    if (a[i] !=  0)
                     {
                    System.out.println(a[i] + " " + count);
                     }
                }
            }

            public static void main(String[] args) 
            {
                Prac_ExeOne score = new Prac_ExeOne();
                score.fillArray(Array);
                score.sort(Array);
                score.count(Array);


            }
        }

输入:

Enter up to 50 nonnegative numbers.
Mark the end of the list with a negative number.
3
3
2
2
-2

输出:

N Count
3 1
2 2
2 1

期望的结果:

简而言之，我希望程序正确计算值，然后在 N 下方左侧显示该值，并在 Count 右侧下方显示其在数组中的次数

Answer 1

对于排序函数，

for (int j = i; j > 1; j--)

应该是

for (int j = i; j > 0; j--)

我把System.out.println(Arrays.toString(Array));放在调用排序之后，并在开头看到一个3，这让我事实上您正在跳过第一个元素。

请注意，还有 way more efficient sorting algorithms .

对于 count 函数，您在错误的时间重置了 count 并且打印过于频繁。我修改如下:

public void count(int[] a)
{
    //for each element in array go through if conditions.
    System.out.println("N " + "Count");
    for (int i = 0; i < a.length - 1; i++)
    {   
        if (a[i] != 0)
        {
            if (a[i] == a[i + 1])
            {
                count++;
            }
            // if the next element is different, we already counted all of the
            //   current element, so print it, then reset the count
            else
            {
                System.out.println(a[i] + " " + count);
                count = 1;
            }
        }
    }
    // we haven't processed the last element yet, so do that
    if (a[a.length-1] != 0)
        System.out.println(a[a.length-1] + " " + count);
}