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Home/ Questions/Q 9185107
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Editorial Team
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Editorial Team
Asked: June 17, 20262026-06-17T19:12:55+00:00 2026-06-17T19:12:55+00:00

I’ve created a Hash-table and I want to remove a node from the linked-list.

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I’ve created a Hash-table and I want to remove a node from the linked-list. The code works for removing the first node but not for removing others.

void intHashTable::remove(int num){
int location = ((unsigned)num) % size;
Node * runner = table[location];
int checker;

if(runner->next == NULL){
    if(num == table[location]->num){
        table[location] = NULL;
    }
}else{
    if(table[location]->num == num){
        table[location] = table[location]->next;
    }else{
        //This part doesn't seem to be working.
        Node *temp = runner->next;
        while(temp != NULL){ 
            if(temp->num == num){
                runner->next = temp->next;
                delete(temp);
                break;
            }
        }
    }
}

}

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-06-17T19:12:56+00:00Added an answer on June 17, 2026 at 7:12 pm

    You haven’t updated temp to point to the next item within the loop:

    temp = temp->next;
    

    You also appear to represent an empty row with a NULL pointer in your table, but you don’t handle this case properly in your code – if runner is NULL then you’ll crash when you try to access runner->next in the first check. Also, you’re failing to delete the node in some cases.

    To fix these issues, you can update your code to something like this:

    void intHashTable::remove(int num)
    {
        int location = ((unsigned)num) % size;
        Node * runner = table[location];
    
        if (runner != NULL) {
            if (runner->num == num) {
                delete runner;
                table[location] = NULL;
            } else {
                while (runner->next != NULL) {
                    if (runner->next->num == num) {
                        Node *temp = runner->next;
                        runner->next = runner->next->next;
                        delete temp;
                        break;
                    }
                    runner = runner->next;
                }
            }
        }
    }
    

    Also note that I’ve removed the brackets from delete, which is a C++ keyword and not a function.

    If you use doubly-linked lists (i.e. with a previous pointer as well as a next) then you can simplify this code a little, although for something like a hash table where you only tend to iterate through in one direction it’s probably not worth the expense of the extra pointer (an extra 8 bytes per item on a 64-bit system).

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