本题代码如下

linklist merge(linklist* L1, linklist* L2)//将两个链表的公共元素合并产生新链表
{
	lnode* ra = (*L1)->next, * rb = (*L2)->next;
	lnode* r;
	lnode* s;
	lnode* C = (lnode*)malloc(sizeof(lnode));
	C->next = NULL;
	r = C;
	while (ra && rb)//循环跳出条件（有一个为空就跳出）
	{
		if (ra->data < rb->data)//若A的当前元素较小，后移指针
			ra = ra->next;
		else if (ra->data > rb->data)//若B的当前元素较小，后移指针
			rb = rb->data;
		else//找到公共结点
		{
			s = (lnode*)malloc(sizeof(lnode));//创建一个结点
			s->data = ra->data;//复制产生的结点*s
			r->next = s;
			r = s;
			ra = ra->next;//A和B继续向后遍历
			rb = rb->next;
		}
	}
	r->next = NULL;//C的表尾结点置空
	return C;
}

完整测试代码

#include<stdio.h>
#include<stdlib.h>
typedef struct lnode
{
	int data;
	struct lnode* next;
}lnode, * linklist;
int n = 5;
int a[5] = { 1,2,3,4,5 };
int b[5] = { 2,4,5,6,7 };
void buildlinklist(linklist* L, int arr[], int n)//创建链表
{
	*L = (lnode*)malloc(sizeof(lnode));
	(*L)->next = NULL;
	lnode* s = *L, * r = *L;
	int i = 0;
	for (i = 0; i < n; i++)
	{
		s = (lnode*)malloc(sizeof(lnode));
		s->data = arr[i];
		s->next = r->next;
		r->next = s;
		r = s;
	}
	r->next = NULL;
}
linklist merge(linklist* L1, linklist* L2)//将两个链表的公共元素合并产生新链表
{
	lnode* ra = (*L1)->next, * rb = (*L2)->next;
	lnode* r;
	lnode* s;
	lnode* C = (lnode*)malloc(sizeof(lnode));
	C->next = NULL;
	r = C;
	while (ra && rb)//循环跳出条件（有一个为空就跳出）
	{
		if (ra->data < rb->data)//若A的当前元素较小，后移指针
			ra = ra->next;
		else if (ra->data > rb->data)//若B的当前元素较小，后移指针
			rb = rb->data;
		else//找到公共结点
		{
			s = (lnode*)malloc(sizeof(lnode));//创建一个结点
			s->data = ra->data;//复制产生的结点*s
			r->next = s;
			r = s;
			ra = ra->next;//A和B继续向后遍历
			rb = rb->next;
		}
	}
	r->next = NULL;//C的表尾结点置空
	return C;
}
void print(linklist* L)//输出单链表
{
	lnode* k = (*L)->next;
	while (k)
	{
		printf("->%d", k->data);
		k = k->next;
	}
}
int main()
{
	linklist L1, L2;
	buildlinklist(&L1, a, n);
	buildlinklist(&L2, b, n);
	printf("L1链表为：");
	print(&L1);
	printf("\nL2链表为：");
	print(&L2);
	linklist C = merge(&L1, &L2);
	printf("\n合并后的链表为：");
	print(&C);
	return 0;
}