《剑指offer》面试题13—O(1)时间删除链表结点

题目：给定单向链表的头指针和某结点指针，实现函数在O(1)时间内删除指定节点。

思路：由于没有要删除结点（j结点）的前一个结点（i结点）指针，通常想法是从头开始遍历找到指定结点的前一个结点（i结点），然后使i结点的指针指向j结点的后一个结点k结点。但是这样是O(n)的时间复杂度，不符合要求。

解决方法很巧妙：由于有要删除的j结点的指针，因此可以很容易得到j结点的后一个结点k结点的指针，只要把k结点的内容复制给i结点，然后删除k结点即可。

但是要注意！如果链表只有一个结点，head指针要指NULL；如果要删除的是尾结点，即后面没有结点了，那么也只能从头开始遍历。

 #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;

 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 DeleteNode(Node* pdelete)
 {
     Node * p;
     if(phead == NULL || pdelete == NULL)   //check
         return;

     if(pdelete == phead)  //the node to be deleted is equal to the head
     {
         if(phead->next == NULL)//only one node
         {
             delete pdelete;
             pdelete = NULL;
             phead = NULL;
         }
         else        //more than one node
         {
             phead = phead->next;
             delete pdelete;
             pdelete = NULL;
         }
     }
     else if(pdelete->next == NULL) //the node to be deleted is the last node
     {
         p = phead;
         while (p->next->next != NULL) p = p->next;  //find the node before the last node
         p->next = NULL;
         delete pdelete;
         pdelete = NULL;
     }
     else //the normal situation,not the head or the tail
     {
         Node* pnext = pdelete->next;
         pdelete->val = pnext->val;
         pdelete->next = pnext->next;
         delete pnext;
         pnext = NULL;
     }
 }

 int main()
 {
     int val,del,flag;
     Node* pdelete;
     cin>>val;
     while(val != )
     {
         AddNode(val);
         cin>>val;
     }
     PrintLink();

     cout<<"Input the node number you want to delete:"<<endl;
     cin>>del;

     pdelete = phead;
     if(phead == NULL) cout<<"No node!"<<endl;

     flag = ;
     for(int i=;i<del&&flag;i++)
     {
         if (pdelete->next!=NULL) pdelete = pdelete->next;
         else
         {
             cout << "The node not exists!"<<endl;
             flag=;
         }
     }
     if(flag) DeleteNode(pdelete);
     PrintLink();

     return ;
 }