【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:http://blog.csdn.net/wtyvhreal/article/details/42474941