计算算术平均值时如何丢弃最大和最小的数字？

Question

我的问题是关于在我的程序计算算术平均值时丢弃最大和最小的数字。

这是我的代码示例：

public static void main(String[] args) {
    int k;
    int r;
    int thelargest;
    int thesmallest;

    Scanner input = new Scanner(System.in);
    System.out.println("Enter the list of number : ");
    String input2 = input.nextLine();

    String[] numbers = input2.split(" ");

    int[] result = new int[numbers.length];
    for (r = 0; r < numbers.length; r++) {
        result[r] = Integer.parseInt(numbers[r]);

    }

    for (k = 0; k < result.length; k++) {
        System.out.print("");
        System.out.println(result[k]);
    }

    System.out.println(" LargestNumber :  " + TheLargestNumber(result));
    System.out.println(" SmallestNumber :  " + TheSmallestNumber(result));
    thelargest = TheLargestNumber(result);
    thesmallest = TheSmallestNumber(result);
    System.out.println("The Arithmetic Mean : " + AirthmeticMean(result));

}

public static int TheSmallestNumber(int[] series) {
    int thesmallest = series[0];
    for (int i = 1; i < series.length; i++) {
        if (series[i] < thesmallest) {

            thesmallest = series[i];
        }
    }
    return thesmallest;
}

public static int TheLargestNumber(int[] series) {
    int thelargest = series[0];
    for (int i = 1; i < series.length; i++) {
        if (series[i] > thelargest) {

            thelargest = series[i];
        }
    }
    return thelargest;
}

public static float AirthmeticMean(int[] result) {
    int sum = 0;
    for (int i = 0; i < result.length; i++) {
        sum += result[i];
    }
    return (float) sum / result.length;
}

我试图找到方法并编写了此示例，但我不知道如何嵌入此代码示例：

         for (int i = 0; i < result.length; i++) {
        if (series[i] != thesmallest && series[i] != thelargest) {
            total = total + seriess[i];
        }
    }

此代码示例对我有帮助吗？

Answer 1

就在你之前

System.out.println("The Arithmetic Mean : " + AirthmeticMean(result));

写

thenewmean = (AirthmeticMean(result)*result.length - thesmallest - thelargest)/(result.length-2)

然后打印thenewmean

System.out.println("The Arithmetic Mean : " + thenewmean);

你不需要写你的

for (int i = 0; i < result.length; i++) {
    if (series[i] != thesmallest && series[i] != thelargest) {
        total = total + seriess[i];
    }
}

在任何地方编码。即使那样，如果您想使用自己的代码， 然后在你的 AirthmeticMean() 函数中使用它

Answer 2

如果删除最高和最小，则需要跟踪计数和总和

public static float AirthmeticMean(int[] result, int theSmallest, int theLargest) {
    int sum = 0;
    int cnt = 0;
    for (int i = 0; i < result.length; i++) {
            if (result[i] != theSmallest && result[i] != theLargest) {
                sum += result[i];
                cnt++;
            }
    }
    return (float) sum / cnt;
}

Answer 3

你得到的代码示例：

for (int i = 0; i < result.length; i++) {
    if (series[i] != thesmallest && series[i] != thelargest) {
        total = total + seriess[i];
    }
}

几乎没问题。几乎，因为您在 for 循环中检索名为 result 的数组的长度并访问数组 series 的元素。

您可以通过以下方式使用此代码。使用两个参数 theSmallest 和 theLargest 扩展 AirthmeticMean 并跟踪求和元素的数量：

public static float AirthmeticMean(int[] result, int theSmallest, int theLargest) {
    int sum = 0;
    int numElements = 0;
    for (int i = 0; i < result.length; i++) {
        if (result[i] != theSmallest && result[i] != theLargest) {
            sum += result[i];
            numElements++;
        }
    }
    return (float) sum / numElements;
}

编辑：添加numElements。