《剑指offer》面试题16—反转链表

Node* p1  p2  p3

思路：开始时，p1为NULL，p2=phead，p3=p2—>next。使p2—>next = p1，然后使p1=p2，p2=p3。如果只有1个结点则此时p2为NULL，结束循环；否则继续执行，直到p2为NULL。

注意：注意代码鲁棒性！判断头结点为NULL；如果只有一个结点。

 #include <iostream>
 using namespace std;

 class Node
 {
 public:
     Node(int v, Node* n)
     {val = v;
     next = n;}
     ~Node(){}
     int val;
     Node* next;
 };
 Node * phead = NULL;
 Node * pReverseHead = NULL;

 void AddNode(int val)
 {
     Node* pnode = new Node(val,NULL);
     //Node one_node(val,NULL); 这里有大bug！如果这样写，在函数退出时自动调用析构函数！链表就乱了！
     if(phead == NULL)
     {
         phead = pnode;
     }
     else
     {
         Node* p = phead;
         while(p->next != NULL)
         {
             p = p->next;
         }
         p->next = pnode;
     }
 }
 void PrintLink()
 {
     Node* p = phead;
     while(p != NULL)
     {
         int temp = p->val;
         cout<<temp<<endl;
         p = p->next;
     }
 }
 void PrintLinkReverse()
 {
     Node* p = pReverseHead;
     while(p != NULL)
     {
         int temp = p->val;
         cout<<temp<<endl;
         p = p->next;
     }
 }
 Node* ReverseLink()
 {
     Node* pPrev = NULL;
     Node* pNode = phead;
     Node* pNext;
     if(phead == NULL) return NULL;
     while(pNode != NULL)
     {
         pNext = pNode->next;
         if(pNext == NULL)
         {
             pReverseHead = pNode;
         }

         pNode->next = pPrev;
         pPrev = pNode;
         pNode = pNext;

     }
     return pReverseHead;
 }
 int main()
 {
     int val;
     cin>>val;
     while(val != )
     {
         AddNode(val);
         cin>>val;
     }
     PrintLink();
     Node* pReverseHead = ReverseLink();
     cout<<"Reverse:"<<endl;
     PrintLinkReverse();
     return ;
 }