剑指offer-面试题5.从尾到头打印链表

题目:输入一个链表的头结点,从尾到头反过来打印出每个结点的值。

 

 

刚看到这道题的小伙伴可能就会想，这还不简单，将链表反转输出。但是这种情况破坏了链表的结构。如果面试官要求不破坏链表结构呢，这时候我们就想到了一种数据结构---栈
 当我们从前往后遍历链表逐个压栈 然后遍历结束后再逐个出栈。

 

首先我们先来用第一种方式实现:

1 #include <iostream>
2 using namespace std;
3
4 struct ListNode
5 {
6     int data;
7     struct ListNode *next;
8 };
9
10 struct ListNode* CreateList()
11 {
12     struct ListNode* Head,*p;
13     Head=(struct ListNode*)malloc(sizeof(ListNode));
14     Head->data=0;
15     Head->next=NULL;
16     p=Head;
17
18     cout<<"Create List....(0-exit!)"<<endl;
19     while(true)
20     {
21         int Data;
22         cin>>Data;
23         if(Data!=0)
24         {
25             struct ListNode* NewNode;
26             NewNode=(struct ListNode*)malloc(sizeof(ListNode));
27             NewNode->data=Data;
28             NewNode->next=NULL;
29             p->next=NewNode;
30             p=p->next;
31         }
32         else
33         {
34             break;
35         }
36     }
37
38     return Head->next;
39 }
40
41 void PrintList(struct ListNode* Head)
42 {
43     cout<<"The List is: ";
44
45     struct ListNode *p;
46     p=Head;
47     while(p!=NULL)
48     {
49         cout<<p->data<<" ";
50         p=p->next;
51     }
52     cout<<endl;
53 }
54
55 struct ListNode* ReversePrint(struct ListNode* Head)
56 {
57     struct ListNode *p1,*p2,*p3;
58
59     p1=Head;
60     p2=p1->next;
61
62     while(p2!=NULL)
63     {
64         p3=p2->next;
65         p2->next=p1;
66         p1=p2;
67         p2=p3;
68     }
69
70     Head->next=NULL;
71     return p1;
72 }
73
74 int main()
75 {
76     ListNode *Head,*NewHead;
77     Head=CreateList();
78     PrintList(Head);
79     NewHead=ReversePrint(Head);
80
81     cout<<endl<<"The Reverse List is:"<<endl;
82     PrintList(NewHead);
83     return 0;
84 }

截图:



 

注意:这种情况下是破坏了链表的结构了。

 

 

下面我们用栈结构来实现不破链表本身结构的逆序输出:

1 #include <iostream>
2 #include <stack>
3 using namespace std;
4
5 struct ListNode
6 {
7     int data;
8     struct ListNode *next;
9 };
10
11 struct ListNode* CreateList()
12 {
13     struct ListNode* Head,*p;
14     Head=(struct ListNode*)malloc(sizeof(ListNode));
15     Head->data=0;
16     Head->next=NULL;
17     p=Head;
18
19     cout<<"Create List....(0-exit!)"<<endl;
20     while(true)
21     {
22         int Data;
23         cin>>Data;
24         if(Data!=0)
25         {
26             struct ListNode* NewNode;
27             NewNode=(struct ListNode*)malloc(sizeof(ListNode));
28             NewNode->data=Data;
29             NewNode->next=NULL;
30             p->next=NewNode;
31             p=p->next;
32         }
33         else
34         {
35             break;
36         }
37     }
38
39     return Head->next;
40 }
41
42 void PrintList(struct ListNode* Head)
43 {
44     cout<<"The List is: ";
45
46     struct ListNode *p;
47     p=Head;
48     while(p!=NULL)
49     {
50         cout<<p->data<<" ";
51         p=p->next;
52     }
53     cout<<endl;
54 }
55
56 void ReversePrintByStack(struct ListNode* Head)
57 {
58     struct ListNode* p;
59     p=Head;
60
61     stack<int> S;
62
63     while(p!=NULL)
64     {
65         S.push(p->data);
66         p=p->next;
67     }
68
69     cout<<"Reverse To Print LinkList: ";
70     while(!S.empty())
71     {
72         cout<<S.top()<<" ";
73         S.pop();
74     }
75
76     cout<<endl;
77 }
78
79 int main()
80 {
81     ListNode *Head,*NewHead;
82     Head=CreateList();
83     PrintList(Head);
84     ReversePrintByStack(Head);
85     return 0;
86 }

截图:



 

哎 好困。