6-1 单链表逆转 (20分)

题目链接:6-1 单链表逆转 (20分)

方式一:递归逆置单链表

 1 #include <stdio.h>
 2 #include <stdlib.h>
 3 #include <math.h>
 4 #include <string.h>
 5 #include <ctype.h>
 6 
 7 #define maxn 5000
 8 #define newline printf("\n")
 9 
10 
11 typedef int ElemType;
12 typedef struct Node *PtrToNode;
13 
14 struct Node{
15     ElemType Data;
16     PtrToNode Next;
17 };
18 
19 typedef PtrToNode List;    //定义单链表类型
20 
21 
22 List Read();
23 void Print(List L);
24 
25 List Reverse(List L);
26 
27 
28 int main(){
29 
30     List L1,L2;
31     L1 = Read();
32     L2 = Reverse(L1);
33     Print(L1);
34     Print(L2);
35     return 0;
36 }
37 
38 
39 List Read(){
40 
41     int n;
42     scanf("%d",&n);
43     int i,temp;
44     List newNode = NULL;
45     List L = (List)malloc(sizeof(struct Node));    //投结点
46     List p = L;
47     if(L!=NULL){
48         for(i=0;i<n;i++){
49             scanf("%d",&temp);
50             newNode = (List)malloc(sizeof(struct Node));
51             newNode->Data = temp;
52             p->Next = newNode;
53             p = p->Next;
54         }
55         p->Next = NULL;
56     }
57     return L->Next;    //根据题意推测返回的是不带头结点的单链表
58     
59 }
60 
61 
62 void Print(List L){
63     List p = L;
64     while(p){
65         printf("%d",p->Data);
66         p = p->Next;
67         if(p!=NULL){
68             printf(" ");
69         }else{
70             printf("\n");
71         }
72     }
73 
74 }
75 
76 
77 List Reverse(List L){
78     
79     if(L == NULL||L->Next == NULL){
80         return L;
81     }
82     List head = Reverse(L->Next);
83     L->Next->Next = L;        //原来的最后一个结点拼接到新链表的最后一个结点后
84     L->Next = NULL;
85 
86     return head;
87 }

方式二:头插法

 1 List Reverse(List L){
 2     
 3     if(L == NULL||L->Next == NULL){
 4         return L;
 5     }
 6     List p = L->Next;
 7     L->Next = NULL;
 8     List head = L;
 9     List q = NULL;
10 
11     while(p){//考察待插入结点p
12         q=p->Next;    //记录下一个待插入结点
13         p->Next = head;    //连接新链表的首元结点
14         head = p;    //成为新链表的新的首元结点
15         p = q;    //新的待插入结点
16     }
17     return head;
18 }

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转载自www.cnblogs.com/ManOK/p/12585365.html
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