【Leetcode -19.删除链表的倒数第N个结点 -24.两两交换链表中的节点】

		struct ListNode* removeNthFromEnd(struct ListNode* head, int n)
		{
		    //创建一个哨兵位，它的next是head
		    struct ListNode* p = malloc(sizeof(struct ListNode));
		    p->val = 0;
		    p->next = head;
		
		    //fast从头结点开始，slow从哨兵位开始
		    //fast和slow拉开n个距离，加上哨兵位，实际上是n+1个距离
		    //这样才可以使要删除的结点的上一个结点直接指向要删除的结点的下一个结点
		    struct ListNode* fast = head, * slow = p;
		    for (int i = 0; i < n; i++)
		    {
		        if (fast == NULL)
		        {
		            return NULL;
		        }
		        fast = fast->next;
		    }
		
		    //然后fast和slow同时走，当fast为空，slow的next就是要删除的结点
		    while (fast)
		    {
		        slow = slow->next;
		        fast = fast->next;
		    }
		
		    //更新slow即可
		    slow->next = slow->next->next;
		
		    //需要返回哨兵位的next，因为如果要删除的结点是头结点，返回头结点就不行
		    struct ListNode* curr = p->next;
		    free(p);
		
		    return curr;
		}

		struct ListNode* swapPairs(struct ListNode* head)
		{
		    //创建一个哨兵位
		    struct ListNode* dummyHead = malloc(sizeof(struct ListNode));
		    dummyHead->val = 0;
		    dummyHead->next = head;
		
		    //curr从哨兵位开始
		    struct ListNode* curr = dummyHead;
		
		    //每次交换的是curr的后两个节点
		    while (curr->next && curr->next->next)
		    {
		        struct ListNode* node1 = curr->next;
		        struct ListNode* node2 = curr->next->next;
		
		        curr->next = node2;
		        node1->next = node2->next;
		        node2->next = node1;
		
		        //更新curr
		        curr = node1;
		    }
		
		    head = dummyHead->next;
		    free(dummyHead);
		    return head;
		}