Remove Linked List Elements

Remove all elements from a linked list of integers that have value val.

Example
Given: 1 --> 2 --> 6 --> 3 --> 4 --> 5 --> 6, val = 6
Return: 1 --> 2 --> 3 --> 4 –> 5

1、头节点特殊处理

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     struct ListNode *next;
 * };
 */
struct ListNode* removeElements(struct ListNode* head, int val) {
    if(head == NULL)
        return NULL;
    struct ListNode* tmp = head;
    struct ListNode* tmp2 = NULL;
    while(head->val == val)
            {
                tmp2 = head;
                head = tmp2->next;
                free(tmp2);
                if(head == NULL)
                    return NULL;
                tmp = head;
            }
    while(tmp->next != NULL)
        {
            if(tmp->next->val == val)
                {
                    tmp2 = tmp->next;
                    tmp->next = tmp2->next;
                    free(tmp2);
                }
            else
                tmp = tmp->next;
        }
    return head;
}

2、使用一个指向head节点的节点来解除特殊处理

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     struct ListNode *next;
 * };
 */
struct ListNode* removeElements(struct ListNode* head, int val) {
    if(head == NULL)
        return NULL;
    struct ListNode* tmp = malloc(sizeof(struct ListNode));
    tmp->next = head;
    head = tmp;
    struct ListNode* tmp2 = NULL;
    while(tmp->next != NULL)
        {
            if(tmp->next->val == val)
                {
                    tmp2 = tmp->next;
                    tmp->next = tmp2->next;
                    free(tmp2);
                }
            else
                tmp = tmp->next;
        }
    return head->next;
}

3、先处理普通情况，最后处理head

struct ListNode* removeElements(struct ListNode* head, int val) {
    if (head == NULL) 
        return NULL;
    struct ListNode *cur = head, *next = head->next;
     while (next != NULL) {
         if (next->val == val) {
             cur->next = next->next;
          next = cur->next;
        } else {
            cur = cur->next;
            next = next->next;
        }
    }
    if (head->val == val) {
         cur = head;
         head = head->next;
     }
     return head;
}

看到最后评分还是不高，不知道更好的方法是什么。。。