Question

我正在研究一个小程序，它计算整数出现在数组中的次数。我设法做到了这一点，但有一件事我无法克服。

我的代码是：

#include <stdio.h>

int count_occur(int a[], int num_elements, int value);
void print_array(int a[], int num_elements);

void main(void)
{
  int a[20] = {2, 5, 0, 5, 5, 66, 3, 78, -4, -56, 2, 66, -4, -4, 2, 0, 66, 17, 17, -4};
  int num_occ, i;

  printf("\nArray:\n");
  print_array(a, 20);

  for (i = 0; i<20; i++)
  {
    num_occ = count_occur(a, 20, a[i]);
    printf("The value %d was found %d times.\n", a[i], num_occ);
  }
}

int count_occur(int a[], int num_elements, int value)
/* checks array a for number of occurrances of value */
{
  int i, count = 0;
  for (i = 0; i<num_elements; i++)
  {
    if (a[i] == value)
    {
        ++count; /* it was found */
    }
  }
  return(count);
}

void print_array(int a[], int num_elements)
{
  int i;
  for (i = 0; i<num_elements; i++)
  {
    printf("%d ", a[i]);
  }
  printf("\n");
}

我的输出是：

Array:
2 5 0 5 5 66 3 78 -4 -56 2 66 -4 -4 2 0 66 17 17 -4 
The value 2 was found 3 times.
The value 5 was found 3 times.
The value 0 was found 2 times.
The value 5 was found 3 times.
The value 5 was found 3 times.
The value 66 was found 3 times.
The value 3 was found 1 times.
The value 78 was found 1 times.
The value -4 was found 4 times.
The value -56 was found 1 times.
The value 2 was found 3 times.
The value 66 was found 3 times.
The value -4 was found 4 times.
The value -4 was found 4 times.
The value 2 was found 3 times.
The value 0 was found 2 times.
The value 66 was found 3 times.
The value 17 was found 2 times.
The value 17 was found 2 times.
The value -4 was found 4 times.

如何避免输出中的双线？

Answer 1

您可以使用并行数组，此示例使用char[20]以节省一些空间：

#include <stdio.h>

int count_occur(int a[], char exists[], int num_elements, int value);
void print_array(int a[], int num_elements);

int main(void) /* int main(void), please */
{
    int a[20] = {2, 5, 0, 5, 5, 66, 3, 78, -4, -56, 2, 66, -4, -4, 2, 0, 66, 17, 17, -4};
    char exists[20] = {0}; /* initialize all elements to 0 */
    int num_occ, i;

    printf("\nArray:\n");
    print_array(a, 20);

    for (i = 0; i < 20; i++)
    {
        num_occ = count_occur(a, exists, 20, a[i]);
        if (num_occ) {
            exists[i] = 1; /* first time, set to 1 */
            printf("The value %d was found %d times.\n", a[i], num_occ);
        }
    }
}

int count_occur(int a[], char exists[], int num_elements, int value)
/* checks array a for number of occurrances of value */
{
    int i, count = 0;

    for (i = 0; i < num_elements; i++)
    {
        if (a[i] == value)
        {
            if (exists[i] != 0) return 0;
            ++count; /* it was found */
        }
    }
    return (count);
}

void print_array(int a[], int num_elements)
{
    int i;
    for (i = 0; i<num_elements; i++)
    {
        printf("%d ", a[i]);
    }
    printf("\n");
}

此方法速度更快，因为它会跳过已经加入的值并开始从i中的count_ocurr进行迭代：

#include <stdio.h>

int count_occur(int a[], char map[], int num_elements, int start);
void print_array(int a[], int num_elements);

int main(void)
{
    int a[20] = {2, 5, 0, 5, 5, 66, 3, 78, -4, -56, 2, 66, -4, -4, 2, 0, 66, 17, 17, -4};
    char map[20] = {0};
    int num_occ, i;

    printf("\nArray:\n");
    print_array(a, 20);

    for (i = 0; i < 20; i++)
    {
        if (map[i] == 0) {
            num_occ = count_occur(a, map, 20, i);
            printf("The value %d was found %d times.\n", a[i], num_occ);
        }
    }
}

int count_occur(int a[], char map[], int num_elements, int start)
/* checks array a for number of occurrances of value */
{
    int i, count = 0, value = a[start];

    for (i = start; i < num_elements; i++)
    {
        if (a[i] == value)
        {
            map[i] = 1;
            ++count; /* it was found */
        }
    }
    return (count);
}

void print_array(int a[], int num_elements)
{
    int i;
    for (i = 0; i< num_elements; i++)
    {
        printf("%d ", a[i]);
    }
    printf("\n");
}

Answer 2

如果当前索引也是相关数字第一次出现的索引，我建议只打印该语句。

在count_occur内，你有i中每个匹配的索引。如果您将i从main传递到count_occur，则可以执行以下操作：如果该值大于i中的count_occur，则返回-1 。然后，如果在main中得到-1，则不要打印。

此外，您的算法可以更快。您可以对数组的副本进行排序，以便有效地完成搜索，而不是每次都线性搜索数组。（即使您使用一个数组进行索引而另一个数组进行搜索，它也会更快 - 并且仍以相同的顺序返回值。）

Answer 3

#include <stdio.h>
#include <stdbool.h>
#include <string.h>

int count_occur(int a[], int num_elements, int value, bool selected[]);
void print_array(int a[], int num_elements);

int main(void){
    int a[] = {2, 5, 0, 5, 5, 66, 3, 78, -4, -56, 2, 66, -4, -4, 2, 0, 66, 17, 17, -4};
    int size = sizeof(a)/sizeof(*a);
    bool ba[size];
    memset(ba, 0, sizeof ba);
    int num_occ, i;

    printf("\nArray:\n");
    print_array(a, size);

    for (i = 0; i<size; i++){
        if(ba[i] == true) continue;//skip already count
        num_occ = count_occur(a, 20, a[i], ba);
        printf("The value %d was found %d times.\n", a[i], num_occ);
    }
}

int count_occur(int a[], int num_elements, int value, bool ba[]){
    int i, count = 0;
    for (i = 0; i<num_elements; i++){
        if (a[i] == value){
            ba[i] = true;
            ++count;
        }
    }
    return count;
}

void print_array(int a[], int num_elements){
    int i;
    for (i = 0; i<num_elements; i++){
        printf("%d ", a[i]);
    }
    printf("\n");
}

改善不大

int count_occur(int a[], int num_elements, int index, bool selected[]);

num_occ = count_occur(a, 20, i, ba);

int count_occur(int a[], int num_elements, int index, bool ba[]){
    int i, count = 0;
    for (i = index; i<num_elements; i++){
        if (a[i] == a[index]){
            ba[i] = true;
            ++count;
        }
    }
    return count;
}

Answer 4

#include<stdio.h>
#include<string.h>

int main()
{
  int arr[] = {2, 5, 0, 5, 5, 66, 3, 78, -4, -56, 2, 66, -4, -4, 2, 0, 66, 17, 17, -4};
  int arrSize = sizeof(arr)/sizeof(arr[0]);
  int tracker[20];
  int i,j,k=0,l=0,count,exists=0;

  for (i=0;i<arrSize;i++)
    printf("%d\t", arr[i]);
  printf("\n");

  memset(tracker, '$', 20);

  for (i=0, j=i+1, count=1, l=0; i<arrSize; i++)
  {
    j=i+1;
    count=1;
    l=0;

    while (l < arrSize)
    {
      if (arr[i] == tracker[l])
      {
        exists = 1;
        break;
      }
      l++;
    }

    if (1 == exists)
    {
      exists = 0;
      continue;
    }

    while (j < arrSize)
    {
      if (arr[i] == arr[j])
        count++;
      j++;
    }
    tracker[k] = arr[i];
    k++;

    printf("count of element %d is %d\n", arr[i], count);
  }

}

Answer 5

very simple logic to count how many time a digit apper
#include<stdio.h>
  int main()
{
int a,b,c,k[10];
int p[10]={0};
int bb[10]={0};
scanf("%d\n",&a);
for(b=0;b<a;b++)
{
    scanf("%d",&k[b]);

}
for(b=a-1;b>0;b--)
{
    for(c=b-1;c>=0;c--)
    {
        if((k[b]==k[c])&&(bb[c]==0))
        {
            p[b]=p[b]+1;
            bb[c]=1;
        }
    }
}
    for(c=0;c<a;c++)
        {
        if(p[c]!=0)
        {
            printf("%d is coming %d times\n",k[c],p[c]+1);
        }
    }
    return 0;
}

Answer 6

在你的功能中：

int count_occur(int a[], int num_elements, int value)
/* checks array a for number of occurrances of value */
{
  int i, count = 0;
  for (i = 0; i<num_elements; i++)
  {
    if (a[i] == value)
    {
        ++count; /* it was found */
        a[i] = INFINITY;              // you can typedef INFINITY with some big number out of your bound
    }
  }  
  return(count);
}

在main（）中你可以编辑for循环：

for (i = 0; i<20; i++)
  {
     if(a[i] != INFINITY)
     {
         num_occ = count_occur(a, 20, a[i]);
         printf("The value %d was found %d times.\n", a[i], num_occ);
     }
  }

计算数字在数组中出现的次数

6 个答案: