链表问题?

[xiansizhe]

/* 大家好,怎样把一个链表分割成两个链表,再把它合成一个循环链表,谢谢大家 */

#include<stdio.h>
#include<stdlib.h>
struct node 
{
	int data;
	struct node *next;
};

struct node *create()  /* 建立链表 */
{
	int num;
	struct node *head, *tail, *p;
	head = tail = NULL;
	puts("Please enter a num: ");
	scanf("%d",&num);
	while(num != 0)
	{
		p = (struct node *)malloc(sizeof(struct node));
		p -> data = num;
		p -> next = NULL;
		if(head == NULL)
			head = tail = p;
		else
		{
			tail -> next = p;
			tail = p;
		}
		scanf("%d",&num);
	}
	return (head);
}

int main(void)
{
	struct node *point,*temp;
	point = create();
    
	puts("The list  ");
	while(point != NULL)  /* 读取链表 */
	{
		temp = point;
		printf("%d\t",point -> data);
		point = point -> next;
		free(temp);
	}
	return 0;
}

[cattycat]

比如从链表的p节点分割，先遍历到p节点，那p的前面的节点的next=NULL，然后返回原来的head和p，就是分割后两个链表的头指针。
合并成循环表，遍历第一个链表，尾节点的next指向head2，然后遍历第二个链表，末尾节点的next=head1.
就行了。

[迷夏牛]

同意楼上的

[bitxinhai]

分割成两个链表和合并都要有一定的规则吧，
不然的话，是很难确定怎么分割的

[星羽]

#include<stdio.h>
#include<stdlib.h>
struct node 
{
    int data;
    struct node *next;
};

struct node *create()  /* 建立链表 */
{
    int num;
    struct node *head, *tail, *p;
    head = tail = NULL;
    puts("Please enter a num: ");
    scanf("%d",&num);
    while(num != 0)
    {
        p = (struct node *)malloc(sizeof(struct node));
        p -> data = num;
        p -> next = NULL;
        if(head == NULL)
            head = tail = p;
        else
        {
            tail -> next = p;
            tail = p;
        }
        scanf("%d",&num);
    }
    return (head);
}

void print_link(struct node* head) 
{
    struct node* point = head;
    puts("The list  ");
    while(point != NULL) 
    {
        printf("%d\t",point -> data);
        point = point -> next;
    }
    printf("\n");
}

void free_link(struct node* head)
{
    struct node* point = head;
    while(point != NULL) 
    {
        struct node* t = point;
        point = point -> next;
        free(t);
    }
}

struct node* split_link(int split_pos, struct node* head)
{
    struct node* point = head;
    struct node* prevp = head;
    struct node* newp = NULL;
    int pos = 0;
    
    if (split_pos <= 0)
        return NULL;
    
    while(point != NULL) 
    {
        if (pos == split_pos)
            break;
        prevp = point;
        point = point -> next;
        ++pos;
    }

    if (point != NULL)
    {
        newp = point;
        prevp->next = NULL; 
    }

    return newp;
}

void merge_link( struct node* p1, struct node* p2)
{
    struct node* point = p1;

    if (!p1 || !p2 || !point)
        return;

    while(point->next != NULL) 
    {
        point = point->next;
    }

    point->next = p2;
}

int main(void)
{
    struct node *point,*point2;
    point = create();

    print_link(point);

    point2 = split_link(3, point);
    print_link(point);
    print_link(point2);

    merge_link(point, point2);
    print_link(point);



    free_link(point);
    return 0;
}


---------

Please enter a num:
1 2 3 4 5 6 7 8 9 0
The list
1       2       3       4       5       6       7       8       9
The list
1       2       3
The list
4       5       6       7       8       9
The list
1       2       3       4       5       6       7       8       9
请按任意键继续. . .

[liaoweixiaoyu]

lz ，你用的是尾插法建表，最好最后有一个判断
struct node *create() /* 建立链表 */
{
int num;
struct node *head, *tail, *p;
head = tail = NULL;
puts("Please enter a num: ");
scanf("%d",&num);
while(num != 0)
{
p = (struct node *)malloc(sizeof(struct node));
p -> data = num;
p -> next = NULL;
if(head == NULL)
head = tail = p;
else
{
tail -> next = p;
tail = p;
}
scanf("%d",&num);
}
if(tail != NULL)
tali->next=NULL; //对于非空链表，表尾置空
return (head);
}

[liaoweixiaoyu]

4楼的合并后的链表是一个循环链表吗？

[luoyong276831622]

同意一楼

[青葱岁月不可追]

up 星羽GG

[kkndciapp]

星羽都已经写好了。。