重写public double average(double x【】)方法，返回数组x【】的元素的算数平均值

以下是重写后的方法：

public double average(double[] x) {
    double sum = 0;
    for (int i = 0; i < x.length; i++) {
        sum += x[i];
    }
    return sum / x.length;
}

该方法接收一个double类型的数组x作为参数，使用for循环遍历数组元素并计算它们的和。最后，将总和除以数组长度即可得到算术平均值。

相关问题

完成下列代码interface ComputerAverage { public double average(double x[]);}class Gymnastics implements ComputerAverage { public double average(double x[]) { int count=x.length; double aver=0,temp=0; for(int i=0;i<count;i++) { for(int j=i;j<count;j++) { if(x[j]<x[i]) { temp=x[j]; x[j]=x[i]; x[i]=temp; } } } for(int i=1;i<count-1;i++) { aver=aver+x[i]; } if(count>2) aver=aver/(count-2); else aver=0; return aver; }}class School implements ComputerAverage { 【代码1】//重写public double average(double x[])方法,返回数组x[]的元素的算术平均}public class Estimator{ public static void main(String args[]) { double a[] = {9.89,9.88,9.99,9.12,9.69,9.76,8.97}; double b[] ={89,56,78,90,100,77,56,45,36,79,98}; ComputerAverage computer; computer=new Gymnastics(); double result= 【代码2】 //computer调用average(double x[])方法，将数组a传递给参数x System.out.print

以下是完成代码的方法：

interface ComputerAverage {
    public double average(double x[]);
}

class Gymnastics implements ComputerAverage {
    public double average(double x[]) {
        int count = x.length;
        double aver = 0, temp = 0;
        for (int i = 0; i < count; i++) {
            for (int j = i; j < count; j++) {
                if (x[j] < x[i]) {
                    temp = x[j];
                    x[j] = x[i];
                    x[i] = temp;
                }
            }
        }
        for (int i = 1; i < count - 1; i++) {
            aver = aver + x[i];
        }
        if (count > 2)
            aver = aver / (count - 2);
        else
            aver = 0;
        return aver;
    }
}

class School implements ComputerAverage {
    //重写public double average(double x[])方法,返回数组x[]的元素的算术平均
    public double average(double x[]) {
        int count = x.length;
        double sum = 0;
        for (int i = 0; i < count; i++) {
            sum += x[i];
        }
        return sum / count;
    }
}

public class Estimator {
    public static void main(String args[]) {
        double a[] = { 9.89, 9.88, 9.99, 9.12, 9.69, 9.76, 8.97 };
        double b[] = { 89, 56, 78, 90, 100, 77, 56, 45, 36, 79, 98 };
        ComputerAverage computer;
        computer = new Gymnastics();
        double result1 = computer.average(a); //computer调用average(double x[])方法，将数组a传递给参数x
        System.out.print("Gymnastics average: " + result1);
        computer = new School();
        double result2 = computer.average(b); //computer调用average(double x[])方法，将数组b传递给参数x
        System.out.print("School average: " + result2);
    }
}

在`School`类中，重写了`average`方法，计算并返回传递给该方法的数组`x[]`的算术平均值。在`Estimator`类的`main`方法中，先使用`Gymnastics`类计算数组`a[]`的平均值，再使用`School`类计算数组`b[]`的平均值，并将结果打印出来。

重写 SolveLinearSystem方法 k 值超出数组A范围

好的，针对这个问题，我们可以在方法内部加上一个简单的判断，如果 k 值超出数组 A 的范围，则返回一个错误提示。下面是代码实现：

public static double[] solveLinearSystem(double[][] A, double[] b) {
    int n = A.length;
    double[] x = new double[n];

    for (int k = 0; k < n; k++) {
        int maxRowIndex = k;
        double maxRowAbsValue = Math.abs(A[k][k]);

        // Find the pivot row
        for (int i = k + 1; i < n; i++) {
            double absValue = Math.abs(A[i][k]);
            if (absValue > maxRowAbsValue) {
                maxRowIndex = i;
                maxRowAbsValue = absValue;
            }
        }

        // Swap the pivot row with the current row (k-th row)
        double[] tmp = A[k];
        A[k] = A[maxRowIndex];
        A[maxRowIndex] = tmp;

        double tmpB = b[k];
        b[k] = b[maxRowIndex];
        b[maxRowIndex] = tmpB;

        // Check if k is within the range of A
        if (k >= A[k].length) {
            System.out.println("Error: k value is out of range!");
            return null;
        }

        // Compute the coefficients and eliminate the k-th variable
        for (int i = k + 1; i < n; i++) {
            double coeff = A[i][k] / A[k][k];
            for (int j = k; j < n; j++) {
                A[i][j] -= coeff * A[k][j];
            }
            b[i] -= coeff * b[k];
        }
    }

    // Backward substitution
    for (int k = n - 1; k >= 0; k--) {
        double sum = 0;
        for (int j = k + 1; j < n; j++) {
            sum += A[k][j] * x[j];
        }
        x[k] = (b[k] - sum) / A[k][k];
    }

    return x;
}

现在，在方法内部会判断 k 是否超出 A 的范围，如果超出，则会打印出一个错误提示，并返回 null。这样，我们就可以避免因为 k 超出 A 范围而导致程序崩溃的问题。