问题一： 链表的反序

 

 1ListNode* reverseList(ListNode* head)
 2{
 3    ListNode *p1, *p2 , *p3;
 4    //链表为空，或是单结点链表直接返回头结点
 5    if (head == NULL || head->next == NULL)
 6         return head;
 7    p1 = head;
 8    p2 = head->next;
 9    while (p2 != NULL)
10    {
11        p3 = p2->next;
12        p2->next = p1;
13        p1 = p2;
14        p2 = p3;
15    }
16    head->next = NULL;
17    head = p1;
18    return head;
19}

问题二： 找出链表的中间元素

 

 1ListNode* find_midlist(ListNode* head)
 2{
 3    ListNode *p1, *p2;
 4    if (head == NULL || head->next == NULL)
 5        return head;
 6    p1 = p2 = head;
 7    while (1)
 8    {
 9        if (p2->next != NULL && p2->next->next != NULL)
10        {
11            p2 = p2->next->next;
12            p1 = p1->next;
13        }
14        else
15            break;
16    }
17    return p1;
18}

问题三：判断一个单链表是否有环

 

 1 int is_looplist (ListNode *head)
 2 {
 3     ListNode *p1, *p2;
 4     p1 = p2 = head;
 5     if (head == NULL || head->next == NULL)
 6         return 0;
 7     while (p2->next != NULL && p2->next->next != NULL)
 8     {
 9         p1 = p1->next;
10         p2 = p2->next->next;
11         if (p1 == p2)
12             return 1;
13     }
14     return 0;
15 }