《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 != 0)
    {
        AddNode(val);
        cin>>val;
    }
    PrintLink();
    Node* pReverseHead = ReverseLink();
    cout<<"Reverse:"<<endl;
    PrintLinkReverse();
    return 0;
}