【Leetcode -206.反转链表 -876.链表的中间结点】

		struct ListNode* reverseList(struct ListNode* head)
		{
		    //空链表返回空
		    if (head == NULL)
		    {
		        return NULL;
		    }
		
		    //定义两个结构体指针newhead，cur
		    struct ListNode* newhead = head, * cur;
		
		    //new找到链表的最后一个结点，最后返回这个结点
		    while (newhead->next)
		    {
		        newhead = newhead->next;
		    }
		    //将尾部赋给cur
		    cur = newhead;
		
		    //head是原链表的头结点，所以反转之后head的next应该是NULL，若head的next不为空循环继续
		    while (head->next)
		    {
		        //每次进来都定义一个结构体指针newtail，从头开始找
		        //直到newtail的next为cur停下
		        struct ListNode* newtail = head;
		        while (newtail->next != cur)
		        {
		            newtail = newtail->next;
		        }
		
		        //将cur的next链接上一个结点
		        cur->next = newtail;
		
		        //如果cur的前一个结点是head，将head的next指向空，即为尾部
		        if (cur->next == head)
		        {
		            head->next = NULL;
		        }
		
		        //每次循环后更新cur
		        cur = newtail;
		    }
		
		    return newhead;
		}

		struct ListNode* reverseList(struct ListNode* head)
		{
		    struct ListNode* prev = NULL, * curr = head;
		
		    //当curr不为空循环继续
		    while (curr)
		    {
		        //next记录下一个结点
		        //prev记录上一个结点
		        struct ListNode* next = curr->next;
		        curr->next = prev;
		        prev = curr;
		        curr = next;
		    }
		
		    return prev;
		}

		struct ListNode* middleNode(struct ListNode* head)
		{
		    //count记录链表的长度
		    //curr用来返回
		    int count = 0;
		    struct ListNode* curr = head;
		
		    //找空指针，统计链表长度
		    while (curr)
		    {
		        curr = curr->next;
		        count++;
		    }
		
		    //让curr返回头部
		    curr = head;
		
		    //中间结点需要走的步长
		    //用循环将curr往后走就行了
		    int mid = count / 2;
		    while (mid)
		    {
		        curr = curr->next;
		        mid--;
		    }
		
		    return curr;
		}

		struct ListNode* middleNode(struct ListNode* head)
		{
		    //两个指针都从头开始
		    struct ListNode* slow = head, * fast = head;
		
		    //长度为偶数时fast为空，为奇数时fast->next为空
		    while (fast && fast->next)
		    {
		        //slow每次走一步
		        //fast每次走两步
		        slow = slow->next;
		        fast = fast->next->next;
		    }
		
		    //最后slow就是要返回的指针
		    return slow;
		}