该题目来自剑指offer的面试题16.

输入一个链表，反转链表后，输出链表的所有元素。

思路如下：定义3个指针，分别指向当前结点，头一个结点，下一个结点，然后通过就地逆置即可。

代码就不太解释了。不理解的，可以参考剑指offer的书，上面写得很详细。

#include 
     
      #include 
      
       typedef struct Node{    int num;    struct Node *next;}NodeHead,*Nodes;void DeleteNode(Nodes head){        if(head==NULL || head->next==NULL)        return;    Nodes pNode,pNext,pPre;    //pPre = head;    pNode = head->next;    pPre = head;    pPre->next = NULL;    while(pNode->next!=NULL){        pNext = pNode->next;        pNode->next = pPre;        pPre = pNode;        pNode = pNext;    }    pNode->next = pPre;    int i ;    for(i = 0 ; i < 8 ;i++)    {        printf("%d",pNode->num);        pNode = pNode->next;    }}void main(){        Nodes head = NULL;     head = (Nodes)malloc(sizeof(NodeHead));    head->next = NULL;    int i ;    for(i = 8 ;i >= 1 ; i--){        Nodes node = (Nodes)malloc(sizeof(NodeHead));        node->num = i;        node->next = head->next;        head->next = node;    }        DeleteNode(head);        /**    Nodes tmp2;    tmp2 = head->next;    for(i = 0 ; i <3 ; i++){        printf("%d",tmp2->num);        tmp2 = tmp2->next;    }    **/            return 0;}