List有序不交叉链表合并

思路清晰，代码自然就美，题目不难。
告诫自己一点，写代码时候脑袋一定要安静，要想清楚。
犯小错误纯粹是瞎折腾，浪费宝贵时间。
View Code
 1 #include<iostream>
 2 using namespace std;
 3 
 4 template<class Type> class List
 5 {
 6 private:
 7     template<class T> class LinkNode
 8     {
 9         public:
10         LinkNode<Type> *link;
11         Type data;
12 
13         LinkNode(LinkNode<Type>* ptr=NULL)
14         {
15             link=ptr;
16         }
17         LinkNode(const Type& item,LinkNode<Type>* ptr=NULL)
18         {
19             data=item;
20             link=ptr;
21         }
22     };
23     LinkNode<Type>* first;
24 
25 public:
26     List()
27     {
28         first=new LinkNode<Type>;
29     }
30 
31     void input(const Type& end)
32     {
33         LinkNode<Type>* curr;
34         Type val;
35         curr=first;
36         cin>>val;
37         while(val!=end)
38         {
39             LinkNode<Type>* newNode=new LinkNode<Type>(val);
40             curr->link=newNode;
41             curr=curr->link;
42             cin>>val;
43         }
44     }
45 
46     void output()
47     {
48         LinkNode<Type>* curr;
49         curr=first;
50         while(curr->link!=NULL)
51         {
52             cout<<curr->link->data<<"";
53             curr=curr->link;
54         }
55         cout<<endl;
56     }
57 
58     LinkNode<Type>* getHead() const
59     {
60         return first;
61     }
62 
63     void mergeNoIntersectList(const List<Type>& list)
64     {
65         LinkNode<Type> *curr1,*curr2,*curr;
66         curr1=first;
67         curr2=list.getHead()->link;   //注意如果getHead()函数不注明是const函数的话那么const对象不能调用非const成员函数
68         while(curr2!=NULL)
69         {
70             if(curr1->link==NULL)  //要插入时一定要从curr1->link判断，用curr判断不行
71             {
72                 curr1->link=curr2;
73                 return;
74             }
75             if(curr2->data<curr1->link->data)
76             {
77                 LinkNode<Type> *newNode=new LinkNode<Type>(curr2->data);
78                 curr=curr1->link;
79                 curr1->link=newNode;
80                 curr1->link->link=curr;
81                 curr2=curr2->link;
82             }
83             curr1=curr1->link;
84         }
85     }
86 };
87 
88 int main()
89 {
90     List<int> list1,list2;
91     list1.input(-1);
92     list2.input(-1);
93     list1.output();
94     list2.output();
95     list1.mergeNoIntersectList(list2);
96     list1.output();
97 
98 }