Remove Duplicates from Sorted List -- LeetCode Solution in C

Given a sorted linked list, delete all nodes that have duplicate numbers, leaving only distinct numbers from the original list.

For example, Given 1->2->3->3->4->4->5, return 1->2->5. Given 1->1->1->2->3, return 2->3.

**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     struct ListNode *next;
 * };
 */
    struct ListNode* deleteDuplicates(struct ListNode* head) {
    }

I have the written the solution for it, but looking for more ideas and suggestions. This solution has passed all the 166 test cases from leetcode.

    /**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     struct ListNode *next;
 * };
 */
struct ListNode* deleteDuplicates(struct ListNode* head) {
 struct ListNode * current = head, *nextNode, *temp;


 //No elements
 if(head == NULL)
 {
     return NULL;
 }

 //Single element linked list
 if(head->next == NULL)
 {
     return head;
 }

     //if the numbers are repeating from the beginning;;
     //then we need to move head;
     while(head !=NULL && head->next !=NULL && head->val == head->next->val){
     if( head->val == head->next->val)
     {
         nextNode = head->next;
         while(nextNode !=NULL  && head->val == nextNode->val){
            temp = nextNode;
        nextNode = nextNode->next;
            free(temp);
         }

         temp = head;
         head = nextNode;
         current = head;
         free(temp);
     }

     }
     //Again a check to see if the list is empty

         if(head == NULL)
         {
             return NULL;
         }


    while(current->next != NULL){

     nextNode = current->next;



        if(nextNode->next !=NULL  && nextNode->val == nextNode->next->val){         

         while(nextNode->next !=NULL  && nextNode->val == nextNode->next->val){
            temp = nextNode->next;
        nextNode->next = temp->next;
         }
        current->next = nextNode->next;

        }else {
            current = nextNode;   
        }

 }

return head;


}