题解 | #反转链表#

非常笨的方法，算上本身，三个指针来回倒腾，因为用了三个指针，特例一大堆。

/**
 * struct ListNode {
 *	int val;
 *	struct ListNode *next;
 * };
 */
/**
 * 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
 *
 * 
 * @param head ListNode类 
 * @return ListNode类
 */
#include <math.h>
struct ListNode* ReverseList(struct ListNode* head ) {
    // write code here
    if(head == NULL)
    {
        return NULL;
    }
    if(head->next != NULL && head->next->next != NULL)
    {
        struct ListNode* head_tmp1 = head->next;//游指针
        struct ListNode* head_tmp2 = head->next->next;//游指针
        if(head_tmp2->next == NULL)
        {
            head->next = NULL;
            head_tmp1->next = head;
            head_tmp2->next = head_tmp1;
            return head_tmp2;
        }
        head->next = NULL;
        while(head_tmp2->next!=NULL)
        {
            head_tmp1->next = head;
            

            head = head_tmp1;
            head_tmp1 = head_tmp2;
            head_tmp2 = head_tmp2->next;
        }
        head_tmp1->next = head;
        head_tmp2->next = head_tmp1;
        return head_tmp2; 
    }
    else if(head->next->next == NULL && head->next != NULL)
    {
        struct ListNode* head_tmp1 = head->next;//游指针
        head->next = NULL;
        head_tmp1->next = head;
        return head_tmp1;
    }
    else 
    {
        return head;
    }
    

}