单链表（快慢指针）

1.删除链表中等于给定值 val 的所有节点。 OJ链接
struct ListNode* removeElements(struct ListNode* head, int val){
    struct ListNode* cur = head;
    struct ListNode* prev = NULL;
    while(cur)
    {
        if(cur->val == val)
        {
            struct ListNode* next = cur->next;
            if(prev == NULL)
                head = next;
            else
                prev->next = next;
            free(cur);
            cur = next;
        }
        else
        {
            prev = cur;
            cur = cur->next;
        }
    }
    return head;
}
2.反转一个单链表。(**头插**） OJ链接
struct ListNode* reverseList(struct ListNode* head){
    struct ListNode* newH, * cur, *next;
    cur = head;
    newH = NULL;
    while(cur)
    {
        next = cur->next;
        cur->next = newH;
        newH = cur;
        cur = next;
    }
    return newH;
}
3.给定一个带有头结点 head 的非空单链表，返回链表的中间结点。如果有两个中间结点，则返回第二个中间结点。（**快慢指针**）OJ链接
struct ListNode* middleNode(struct ListNode* head){
    struct ListNode* fast,*slow;
    fast = slow = head;
    while(fast && fast->next)
    {
        slow = slow->next;
        fast = fast->next->next;
    }
    return slow;
}
4.输入一个链表，输出该链表中倒数第k个结点。(**快慢指针**) OJ链接
ListNode* FindKthToTail(ListNode* pListHead, unsigned int k) {
  ListNode* slow, *fast;
  fast = slow = pListHead;
    while(k--)
    {
      if(fast）
        fast = fast->next;
      else
        return NULL;
    }
    while(fast)
    {
      slow = slow->next;
      fast = fast->next;
    }
    return slow;
  }