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。我需要一个函数来找到中位数，然后打印出来。此外，我的平均值只显示整数，没有数字，就像我使用浮点数时所认为的那样。但找到一个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，并对其进行四舍五入，然后将其转换为浮点数并分配给avg。您可以通过先将值转换为浮点数，然后执行除法来轻松解决此问题：

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

至于中位数，由于输入数组是排序的，所以只要找到长度并索引中间值即可。如果长度是偶数，取两个中间值的平均值。

Answer 2

首先将序列从大到小lisin然后count / 2通过获取索引元素应该 int medianNumber=count/2; int median=number[medianNumber];

Answer 3

我发现了为什么平均没有工作，我在制作函数时使用了 int average 而不是 float average。我还发现了如何为中值创建函数。完整代码如下

#include <stdio.h>
#include <stdlib.h>

#define NUMBERS_SIZE 10

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;
}

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

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

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

return avg;
}

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


float median(int numbers[] , int count)
{
float medi=0;

// if number of elements are even
if(count%2 == 0)
    medi = (numbers[(count-1)/2] + numbers[count/2])/2.0;
    // if number of elements are odd
else
    medi = numbers[count/2];

return medi;
}


int main()
{
int numbers[NUMBERS_SIZE];

int n;

float medi=0;


int count = sizeof(numbers)/ sizeof(numbers[0]);

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










printf("Minimum: %d\n", minimum(numbers, count));

printf("Maximum: %d\n", maximum(numbers, count));

printf("Sum: %i\n", sum(numbers, count));


printf("Average: %g\n", average(numbers, count));



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

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



medi = median(numbers , count);

printf("\nMedian: %f\n", medi);

return 0;

}