删除 C 中数组中元素的出现

Question

This C code is for deleting all occurrences of an integer in an array. However, when I executed it, there is a problem with displaying the final array, the code doesn't display the rest of the array once it finds the first occurrence.

unsigned int T[10], n, i, j, exist, integerDeleteOccurences;

printf("Array length : ");
scanf("%u", &n);

for(i=0; i<n; i++)
{
    printf("%u | ", T[i]);
}

printf("The number you want to delete its occurences : ");
scanf("%u", &integerDeleteOccurences);

exist = 0;

for (i=0; i<n; i++)
{
    if (T[i] == integerDeleteOccurences)
    {
        j = i;
        for (j=i; j<n-1; j++);
        {
            T[j] = T[j+1];
        }
        exist = 1;
        i--;
        n--;
    }
}

if (exist == 1)
{
    for (i=0; i<n; i++)
    {
        printf("%u | ", T[i]);
    }
}

else if (exist == 0)
{
    printf("This number doesn't exist in the array ! \n");
}

Answer 1

When working with statically allocated arrays (ie you know the maximum possible size ), you should handle them by keeping track of their current size.

Here's a function that delete all occurrencies of an element, given an array and its size, and returns the number of deletions:

int deleteAllOccurrencies(int* arr, int size, int el)
{
    int occurrencies = 0;

    for (int i = 0; i < size; i++)
    {
        if (arr[i] == el)
        {
            occurrencies++;
            
            // shift following elements
            for (int j = i; j < size; j++)
            {
                arr[j] = arr[j + 1];
            }
        }
    }
    return occurrencies;
}

Edit with alternative solution (suggested by chqrlie )

The above function loops through an array of integers and for each occurrency found, removes the element from the array and shifts the following values by one position. However, that is not much efficient, since the time complexity of that approach is O(n²) .

A better solution would be to loop through the array by using two indexes:

• i , which is used to check each value in the starting array, and is increased at the end of each loop;
• j , which is used to update only the array elements that are different from the one we are checking, and is increased only when that value is different.

This way we are able to get a much more efficient check, reaching a time complexity of O(n) :

int deleteAllOccurrencies(int* arr, int size, int el)
{
    int occurrencies = 0;

    for (int i = 0, j = 0; i < size; i++)
    {
        if (arr[i] == el)
        {
            occurrencies++;
        }
        else
        {
            arr[j++] = arr[i];
        }
    }
    return occurrencies;
}

Example Usage

#include <stdio.h>
#define MAX_SIZE 10

int deleteAllOccurrencies(int* arr, int size, int el);
void printArray(int* arr, int size);

int main(int argc, char** argv)
{
    int array[MAX_SIZE] = { 1, 2, 3, 4, 2, 6, 7, 8, 2, 10 };
    int size = MAX_SIZE, res;

    printf("Array: ");
    printArray(array, size);

    res = deleteAllOccurrencies(array, size, 2);
    size = MAX_SIZE - res;

    printf("\nResult: %d occurrencies found!\n", res);
    printf("Resulting array: ");
    printArray(array, size);

    return 0;
}

int deleteAllOccurrencies(int* arr, int size, int el)
{
    int occurrencies = 0;

    for (int i = 0, j = 0; i < size; i++)
    {
        if (arr[i] == el)
        {
            occurrencies++;
        }
        else
        {
            arr[j++] = arr[i];
        }
    }
    return occurrencies;
}

void printArray(int* arr, int size)
{
    printf("[ ");
    for (int i = 0; i < size; i++)
    {
        printf("%d", arr[i]);
        if (i < size - 1)
            printf(", ");
    }
    printf("]\n");
}

Answer 2

It is far to complicated.

size_t removeFromArray(int *arr, size_t size, int val)
{
    int *tail = arr;
    size_t newSize = size;
    if(arr)
    {
        while(size--)
        {
            if(*tail == val) { tail++; newSize--;}
            else 
                *arr++ = *tail++;
        }
    }
    return newSize;
}