这个删除链表中所有x要怎么实现 我写完之后这个for循环在第二三种情况下停不下来

Question

int DeleteNodes(Node** phead, double x) {
    Node* prev = NULL;
    Node*** pphead = &phead;
    int count = 0;
    for (int pos = 1; currNode; prev = currNode, currNode = currNode->next) {
        if (currNode->data == x) {
            if (prev) {
                prev->next = currNode->next;
                free(currNode);
                count++;
                currNode = prev;

            }
            else {
                *phead = currNode->next;
                free(currNode);
                count++;
                currNode = **pphead;
                if (currNode->next == NULL)
                    free(currNode);
            }
        }
    }
    return count;
}*/

Answer 1

 
int DeleteNodes(Node** phead, double x) {
        int count = 0;
        Node* current =*phead;
        Node* pre=NULL;
        while (current != NULL) {
            if (current->data == x) {
                if (pre == NULL) {
                        //如果删除的节点是头结点，那么原来链表的第二个节点就是头节点，此时pre依旧是NULL
                    *phead = current->next;
                    count++;
                     free(current);
                     current = *phead;
                }
                else {
                        //删除的节点不是头结点，那么pre的下一个节点就是当前节点的下一个节点
                    pre->next = current->next;
                    count++;
                    free(current);
                    current = pre->next;
                }
            }
            else {
                pre = current;
                current = current->next;
            }
           
        }
        return count;
}