c - 使用 c 中的函数从排序数组中查找中位数

Question

我有一个程序，它从数组中获取整数并打印出最大、最小、平均值和排序列表。我需要一个函数的帮助来找到中位数，然后将其打印出来。另外，我的平均值只显示整数，没有数字，正如我在使用 float 时所想的那样。但找到中值函数是重要的部分。

#include <stdio.h>

#define NUMBERS_SIZE 5

int sum(int numbers[], int count)
{

    int sum = 0;

    for (int i = 0; i < count; i++)
        sum += numbers[i];

    return sum;
}


int maximum(int numbers[], int count)
{
    int a;


    int max = numbers[0];


    for (a = 1; a < count; a++)
        if (numbers[a] > max)
            max = numbers[a];

    return max;
}

int minimum(int numbers[], int count)
{
    int a;


    int min = numbers[0];


    for (a = 1; a < count; a++)
        if (numbers[a] < min)
            min = numbers[a];

    return min;
}

int average(int numbers[], int count)
{
    float avg;
    int sum = 0;

    for (int i = 0; i < count; i++)
        sum += numbers[i];

    avg = sum/count;

    return avg;
}

int cmpfunc (const void * a, const void * b) {
    return ( *(int*)a - *(int*)b );
}




int main()
{
    int numbers[NUMBERS_SIZE];

    int n;

    float median=0;

    for (int i = 0; i < NUMBERS_SIZE; i++)
    {
        printf("Enter integer: ");
        scanf("%i", &numbers[i]);
    }

    int result3 = sum(numbers, sizeof(numbers) / sizeof(numbers[0]));
    printf("The sum is: %i\n", result3);



    int count3 = sizeof(numbers)/sizeof(numbers[0]);
    printf("Largest number entered is: %d\n", maximum(numbers, count3));

    int count1 = sizeof(numbers)/sizeof(numbers[0]);
    printf("Smallest number entered is: %d\n", minimum(numbers, count1));

    int count4 = sizeof(numbers)/sizeof(numbers[0]);
    printf("Average is %d\n", average(numbers, count4));



    qsort(numbers, NUMBERS_SIZE, sizeof(int), cmpfunc);

    printf("\nSorted: \n");
    for( n = 0 ; n < NUMBERS_SIZE; n++ ) {
        printf("%d ", numbers[n]);
    }


    return 0;

}

Answer 1

您的平均值四舍五入为整数值，因为 sum 和 count 都是整数。然后，在 avg = sum/count 行上，它首先计算 sum/count，四舍五入，然后将其转换为 float 并分配给 avg。您可以通过先将值转换为 float ，然后执行除法来轻松解决此问题:

avg = (float) sum / (float) count;

至于中位数，由于输入数组已排序，因此只需找到长度并索引中间值即可。如果长度是偶数，则取中间两个值的平均值。