【注意】只需要完成该函数功能算法,中间不需要有任何IO 的输入输出
// 返回新链表头节点
LinkNode *reverse_link(LinkNode *head)
{
if(head == NULL)
return NULL;
LinkNode *prev , *curr , *reverse_head , *temp;
prev = NULL , curr = head;
while(curr->next)
{
temp = curr->next;
curr->next = prev;
prev = curr;
curr = temp;
}
curr->next = prev;
reverse_head = curr;
return reverse_head;
} 原文地址:http://blog.csdn.net/wtyvhreal/article/details/42474941