单链表-从中间删除

Question

I am trying to figure out an algorithm to delete from the middle of a linked list..

My idea is to traverse the list, find the node right before the node I want to delete, call it Nprev, and set Nprev to Nnext where Nnext is after the node to delete Ndelete.

So Nprev -> Ndelte -> Nnext .

My problem is that I cannot figure out how to traverse this list to find the node before the one I want.

I've been doing this with seg faults because I assign pointers out of range I assume. Its a very messy algorithm that I have, with many if else statements..

Is there an easier way to do this?

Basically I need to go through the list, apply a function to each node to test if it is true or false. If false I delete the node. Deleting first and last is not as hard but middle stumped me.

Please let me know if there are some general ways to solve this problem. I've been scouring the internet and found nothing I need.

I used this: http://www.cs.bu.edu/teaching/c/linked-list/delete/

but the algorithm before step 4 only deletes the first node in my list and doesn't do any more. How can I modify this?

They also give a recursive example but I don't understand it and am intimidated by it.

Answer 1

First you need to find the middle node. Well take 3 pointers fast, slow, prev with fast moving with twice the speed of slow and prev storing the address of the node previous of slow. ie *slow=&head,*fast=&head,prev=Null traverse the list and when fast=NULL slow will point to the middle node if number of elements are odd and prev will store the address of node previous of the mid node. so simply prev->next=slow->next .

Answer 2

Here an example of something I use to search and remove by index:

Given this struct : (Can also be adapted to other self referencing structs)

struct node
{
    S s;
    int num;
    char string[10];
    struct node *ptr;
};
typedef struct node NODE;

Use this to remove an item from somewhere in the "middle" of the list (by index)

int remove_by_index(NODE **head, int n) /// tested, works 
{
    int i = 0;
    int retval = -1;
    NODE * current = *head;
    NODE * temp_node = NULL;

    if (n == 0) {
        return pop(head);
    }

    for (int i = 0; i < n-1; i++) {
        if (current->ptr == NULL) {
            return -1;
        }
        current = current->ptr;
    }

    temp_node = current->ptr;
    retval = temp_node->num;
    current->ptr = temp_node->ptr;
    free(temp_node);

    return retval;

}