C语言程序设计之链表篇2

链表

        考查链表的数据结构，需利用指针变量才能实现，一个结点中应包含一个指针变量，用来存放下一个结点的地址。
        建立单向链表的一般步骤是：建立头指针 -> 建立第一个结点 -> 头指针指向第一个结点 -> 建立第二个结点 -> 头指针指向第二个结点
-> ··· -> 最后一个结点的指针指向 $NULL$。

问题2_1

        函数 $fun$ 的功能是：将不带头结点的单向链表逆置，即若原链表中从头自尾结点数据域依次为 $2、4、6、8、10$ ，逆置后，依次为 $10、8、6、4、2$ 。

代码2_1

#include<stdio.h>
#include<stdlib.h>

#define N 5


typedef struct node{
	int data;
	struct node *next;
}NODE;

NODE *fun(NODE *h){
	NODE *p, *q, *r;
	p = h;
	if(p==NULL)
		return NULL;
	q = p->next;
	p->next = NULL;
	while(q){
		r = q->next;
		q->next = p;
		p = q;
		q = r;
	}
	return p;
} 

NODE *creatlist(int a[]){
	NODE *h, *p, *q;
	int i;
	h = NULL;
	for(i=0; i<N; i++){
		q = (NODE*)malloc(sizeof(NODE));
		q->data = a[i];
		q->next = NULL;
		if(h==NULL)
			h = p = q;
		else{
			p->next = q;
			p = q;
		}
	}
	return h;
}

void outlist(NODE *h){
	NODE *p;
	p = h;
	if(p==NULL)
		printf("The list is NULL!\n");
	else{
		printf("\nHead");
		do{
			printf("->%d", p->data);
			p = p->next;
		}while(p!=NULL);
		printf("->End\n");
	}
}

void main(void){
	NODE *head;
	int a[N] = {2, 4, 6, 8, 10};
	head = creatlist(a);
	printf("\nThe original list:\n");
	outlist(head);
	head = fun(head);
	printf("\nThe list after inverting:\n");
	outlist(head);
}
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70

结果2_1

问题2_2

        函数 $fun$ 的功能是：将不带头结点的单向链表结点数据中的数据从小到大排序。即若原链表中从头自尾结点数据域依次为 $10、4、2、8、6$ ，排序后，依次为 $2、4、6、8、10$ 。

代码2_2

#include<stdio.h>
#include<stdlib.h>

#define N 6
typedef struct node{
	int data;
	struct node *next;
}NODE;

void fun(NODE *h){
	NODE *p, *q;
	int t;
	p = h;
	while(p){
		q = p->next;
		while(q){
			if(p->data>q->data){
				t = p->data;
				p->data = q->data;
				q->data = t;
			}
			q = q->next;
		}
		p = p->next;
	}
} 

NODE *creatlist(int a[]){
	NODE *h, *p, *q;
	int i;
	h = NULL;
	for(i=0; i<N; i++){
		q = (NODE*)malloc(sizeof(NODE));
		q->data = a[i];
		q->next = NULL;
		if(h==NULL) 
			h = p = q;
		else{
			p->next = q;
			p = q;
		}
	}
	return h;
}

void outlist(NODE *h){
	NODE *p;
	p = h;
	if(p==NULL)
		printf("The list is NULL!\n");
	else{
		printf("\nHead");
		do{
			printf("->%d", p->data);
			p = p->next;
		}
		while(p!=NULL);
		printf("->End\n");
	}
}

void main(void){
	NODE *head;
	int a[N] = {0, 10, 4, 2, 8, 6};
	head = creatlist(a);
	printf("\nThe original list:\n");
	outlist(head);
	fun(head);
	printf("\nThe list after inverting :\n");
	outlist(head);
}
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71

结果2_2

问题2_3

        在带头结点的单向链表中，查找数据域中值为 $ch$ 的结点。找到后通过函数值返回该结点在链表中所处的顺序号；若不存在值为 $ch$ 的结点，函数返回 0 值。

代码2_3

#include<stdio.h>
#include<stdlib.h>

#define N 8

typedef struct list{
	int data;
	struct list *next;
}SLIST;

SLIST *creatlist(char *);
void ooutlist(SLIST *);

int fun(SLIST *h, char ch){
	SLIST *p;
	int n=0;
	p = h->next;
	while(p!=NULL){
		n++;
		if(p->data==ch)
			return n;
		else
			p = p->next;
	}
	return 0;
}

void main(void){
	SLIST *head;
	int k;
	char ch;
	char a[N] = {'z', 'y', 's', 'd', 'a', 'o'};
	head = creatlist(a);
	outlist(head);
	printf("Enter a letter: ");
	scanf("%c", &ch);
	k = fun(head, ch);
	if(k==0)
		printf("\nNot found!\n");
	else
		printf("The sequence number is : %d\n", k);
}

SLIST *creatlist(char *a){
	SLIST *h, *p, *q;
	int i;
	h = p = (SLIST*)malloc(sizeof(SLIST));
	for(i=0; i<N; i++){
		q = (SLIST*)malloc(sizeof(SLIST));
		q->data = a[i];
		p->next = q;
		p = q;
	}
	p->next = 0;
	return h;
}

void outlist(SLIST *h){
	SLIST *p;
	p = h->next;
	if(p==NULL)
		printf("\nThe list is NULL!\n");
	else{
		printf("\nHead");
		do{
			printf("->%c", p->data);
			p = p->next;
		}while(p!=NULL);
	}
	printf("->End\n");
}
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71

结果2_3

问题2_4

         函数 $fun$ 的功能是 ：统计带头结点的单向链表中结点的个数，并存放在形参 $n$ 所指的存储单元中。

代码2_4

#include<stdio.h>
#include<stdlib.h>

#define N 8

typedef struct list{
	int data;
	struct list *next;
}SLIST;

SLIST *creatlist(int *);
void ooutlist(SLIST *);

int fun(SLIST *h, int *n){
	SLIST *p;
	p = h->next;
	while(p!=NULL){
		(*n)++;
		p = p->next;
	}
	return 0;
}

void main(void){
	SLIST *head;
	int a[N] = {12, 87, 45, 32, 91, 16, 20, 48}, num;
	head = creatlist(a);
	outlist(head);
	fun(head, &num);
	printf("\nNumber = %d\n", num);
}

SLIST *creatlist(int *a){
	SLIST *h, *p, *q;
	int i;
	h = p = (SLIST*)malloc(sizeof(SLIST));
	for(i=0; i<N; i++){
		q = (SLIST*)malloc(sizeof(SLIST));
		q->data = a[i];
		p->next = q;
		p = q;
	}
	p->next = 0;
	return h;
}

void outlist(SLIST *h){
	SLIST *p;
	p = h->next;
	if(p==NULL)
		printf("\nThe list is NULL!\n");
	else{
		printf("\nHead");
		do{
			printf("->%d", p->data);
			p = p->next;
		}while(p!=NULL);
	}
	printf("->End\n");
}
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60

结果2_4

问题2_5

         函数 $fun$ 的功能是 ：将链单向链表结点（不包括头结点）数据为偶数的值累加起来，并且作为函数值返回。

代码2_5

#include<stdio.h>
#include<conio.h>
#include<stdlib.h>

typedef struct aa{
	int data;
	struct aa *next;
}NODE;

int fun(NODE * h){
	int sum=0;
	NODE *p;
	p = h->next;
	while(p){
		if(p->data%2==0)
			sum += p->data;
		p = p->next;
	}
	return sum;
}

NODE *creatlink(int n){
	NODE *h, *p, *s;
	int i;
	h = p = (NODE*)malloc(sizeof(NODE));
	for(i=1; i<n; i++){
		s = (NODE*)malloc(sizeof(NODE));
		s->data = rand()%16;
		s->next = p->next;
		p->next = s;
		p = p->next; 
	}
	p->next = NULL;
	return h;	
}

outlink(NODE *h){
	NODE *p;
	p = h->next;
	printf("\n\nThe list : \n\n HEAD");
	while(p){
		printf("->%d", p->data);
		p = p->next;
	}
	printf("\n");
}

void main(void){
	NODE *head;
	int sum;
	head = creatlink(10);
	outlink(head);
	sum = fun(head);
	printf("\nSum = %d", sum);
}
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55

结果2_5