反序排列一个单链表--C和指针，12.8.4

主要思路是：

1.计算链表的节点

2.从而为一个指针数组分配内存，每个元素都是一个指向节点的指针，例如 NODE * temp[3] = {NODE*p1,NODE*p2,NODE*p3};

3. 将每个结构赋给上述的指针

4.使当前结构内的指针成员，例如p3->fwd = p2, p2->fwd = p1 ....

 

 1 # include <stdio.h>
 2 # include <stdlib.h>
 3 
 4 struct NODE {
 5         int value;
 6         struct NODE * fwd;
 7 }Node;
 8 struct NODE* sll_reverse(struct NODE* first);
 9 
10 int main()
11 {
12     NODE a, b, c;

13     c = { 15,NULL };
14     b = { 10, &c };
15     a = { 5, &b };

16     NODE* first = &a;

17     while (first !=NULL)             /打印原先的链表内各节点的value
18     {
19         printf("%d ", first->value);
20         first = first->fwd;
21     }
22     putchar('\n');
23     
24     first = sll_reverse(&a);
25 
26     
27     while (first)                 //重新排序后
28     {
29         printf("%d ", first->value);
30         first = first->fwd;
31     }
32 
33     return 0;
34 
35 }
36 struct NODE* sll_reverse(struct NODE* first)
37 {
38     NODE* temp = first;
39     if (first == NULL)                   //空链表
40         return NULL;
41 
42     int count = 0;
43     while (temp)                 //节点计数
44     {
45         count += 1;
46         temp = temp->fwd;
47     }
48 
49     NODE**reverse_node = (NODE**)malloc(sizeof(NODE*) * count);   //指针数组
50     if (reverse_node == NULL)
51     {
52         printf("Fail to allocate memeory");
53         exit(1);
54     }
55 
56     reverse_node[0] = first;                //为数组的元素分配节点
57     for(int i = 1; i< count;i++)             
58     {
59         reverse_node[i] = first->fwd;
60         first = first->fwd;
61     }
62 
63     first = reverse_node[count-1];                   //改变节点内指针的对象
64     while ((count -2)>=0)
65     {
66         reverse_node[count-1]->fwd = reverse_node[count - 2];
67         count--;
68     }
69     reverse_node[0]->fwd = NULL;                   //尾部节点内的指针声明为NULL
70        
71     free(reverse_node);
72 
73     return first;
74 
75 }

 

posted @ 2022-07-21 17:02  Tolerieren  阅读(27)  评论(0)    收藏  举报