C语言删除链表的倒数第二个元素，补全一下这个代码

Question

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

struct node {
struct node *next;
int data;
};

struct node *delete_second_last(struct node *head);
struct node *strings_to_list(int len, char *strings[]);
void print_list(struct node *head);

// DO NOT CHANGE THIS MAIN FUNCTION

int main(int argc, char *argv[]) {
// create linked list from command line arguments
struct node *head = strings_to_list(argc - 1, &argv[1]);

struct node *new_head = delete_second_last(head);
print_list(new_head);

return 0;

}

//
// Delete the second last node in the list.
// The deleted node is freed.
// The head of the list is returned.
//
struct node *delete_second_last(struct node *head) {

// PUT YOUR CODE HERE (change the next line!)
return NULL;

}

// DO NOT CHANGE THIS FUNCTION
// create linked list from array of strings
struct node *strings_to_list(int len, char *strings[]) {
struct node *head = NULL;
for (int i = len - 1; i >= 0; i = i - 1) {
struct node *n = malloc(sizeof (struct node));
assert(n != NULL);
n->next = head;
n->data = atoi(strings[i]);
head = n;
}
return head;
}

// DO NOT CHANGE THIS FUNCTION
// print linked list
void print_list(struct node *head) {
printf("[");

for (struct node *n = head; n != NULL; n = n->next) {
    // If you're getting an error here,
    // you have returned an invalid list
    printf("%d", n->data);
    if (n->next != NULL) {
        printf(", ");
    }
}
printf("]\n");

}

Answer 1

struct node *delete_second_last(struct node *head) {
    struct node *p = head;
    if(p == NULL)
        return head;
    struct node *q  = p->next;
    if(q == NULL)
      return head;
    if(q->next == NULL)
   {
        free(p);
        return q;
   }
   struct node *m  = q->next;
   while(m != NULL)
   {
        p = q;
        q = m;
        m = m->next;
        if(m == NULL)
        {
              p->next = m;
              free(q);
              break;
        }
    }
    return head;
}