要解决这个问题：
我们可以使用方法重载来捕获所有数据
每个方法将使用不同的数据类型，但名称相同-data（）
应该找出每个数组的空值数。
变量n将确定3个整数中最大的一个。
n将是打印表时的测试表达式限制

public class overLoadEg {
//array that will store integers
    static int[] intArray = new int[10];
//array that will store doubles
    static double[] doubleArray = new double[10];
//array that will store strings
    static String[] stringArray = new String[10];
    static int i = 0, j = 0, k = 0, m, n;

    public static void main(String[] args) {
//input values
        data(23);
        data(23.4554);
        data("Hello");
        data("world");
        data("help");
        data(2355);
        data(52.56);
        data("val");
        data("kkj");
        data(34);
        data(3);
        data(2);
        data(4);
        data(5);
        data(6);
        data(7);
        data(8);
        display();
    }

    public static void data(int val){
        //add int value to int array
        intArray[i] = val;
        System.out.println("Int " + intArray[i] + " added to IntArray");
        i++;
    }
    public static void data(Double val){
        //add double value to double array
        doubleArray[j] = val;
        System.out.println("Double " + doubleArray[j] + " added to doubleArray");
        j++;
    }
    public static void data(String val){
        //add string value to stringarray
        stringArray[k] = val;
        System.out.println("String " + stringArray[k] + " added to stringArray");
        k++;
    }

    public static void max(){
        //To get the maximum number of values in each array
        int x, y, z;
        x = y = z = 0;

        //counting all the null values in each array and storing in x, y and z
        for(m=0;m<10;m++){
            if(intArray[m] == 0){
                ++x; 
            }
            if(doubleArray[m] == 0){
                ++y;
            }
            if(stringArray[m] == null){
                ++z;
            }
        }

        //subtracting the null/0 count from the array size
        //this gives the active number of values in each array
        x = 10 - x;
        y = 10 - y;
        z = 10 - z;

        //comparing all 3 arrays and check which has the max number of values
        //the max numbe is stored in n
        if(x > y){
            if(x > z){
                n = x;
            }
            else{
                n = z;
            }
        }
        else{
            if(y > z){
                n = y;
            }
            else{
                n = z;
            }
        }

    }

    public static void display(){
        //printing the arrays in table
        //All the null/0 values are excluded

        System.out.println("\n\nInt\tDouble\t\tString");
        max();
        for(m = 0; m < n; m++){
            System.out.println(intArray[m] + "\t" + doubleArray[m] + "\t\t" + stringArray[m]);
        }
        System.out.println("Count : " + m);
    }

}