怎么用链表实现两个线性表的合并

如何用链表实现两个线性表的合并
LA=(3,5,8,11)
LB=(2,6,8,9,11,15,20)
相同元素不保留
预测输出：LA=(3,5,8,11,2,6,9,15,20)

[解决办法]
自己加一个删除相同节点的函数，再连接就行

C/C++ code

#include <stdio.h>#include <stdlib.h>#define MAX1 10#define MAX2 6typedef struct list{   int num;   struct list *next;}*link;int data1[MAX1] = { 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 };int data2[MAX2] = { 11, 12, 13, 14, 15, 16 };link create_list(link head, int *data, int MAX){   link newnode;   link pointer;   int i;   head = malloc(sizeof(*head));   if(head==NULL)      printf("Memory allocate Failure!\n");   else   {      head->num = data[0];      head->next = NULL;      pointer = head;      for(i=1; i<MAX; ++i)      {         newnode = malloc(sizeof(*newnode));         if(newnode==NULL) break;         newnode->num = data[i];         newnode->next = NULL;         pointer->next = newnode;         pointer = newnode;      }   }   return head;}void free_list(link head){   link pointer;   while(head!=NULL)   {      pointer = head;      head = head->next;      free(pointer);   }}void print_list(link head){   if(head==NULL)      printf("empty!");   while(head!=NULL)   {      printf("[%d]", head->num);      head = head->next;   }   putchar('\n');}link concat(link head1, link head2){   link pointer = head1;   if(head1!=NULL && head2!=NULL)   {      while(pointer->next!=NULL)         pointer = pointer->next;      pointer->next = head2;   }   else if(head1==NULL && head2!=NULL)      head1 = head2;   return head1;}int main(void){   link head  = NULL;   link head1 = NULL;   link head2 = NULL;      head1 = create_list(head1, data1, MAX1);   head2 = create_list(head2, data2, MAX2);    if(head1!=NULL && head2!=NULL)   {      printf("Input data :\n");      print_list(head1);      print_list(head2);      head = concat(head1, head2);      //head = concat(NULL, head2);       //head = concat(head1, NULL);      //head = concat(NULL, NULL); free(head1); free(head2);      printf("After concat:\n");      print_list(head);      free(head1);   }   return 0;}
[解决办法]
C/C++ code
根据你的要求写了一个#include <stdio.h>#include <malloc.h>typedef struct NODE{    int data;    struct NODE *next;}LNode,*Linklist;void creatList(Linklist L,int LA[],int len);int judgeList(Linklist L,int x);int insertList(Linklist L,int x);void printList(Linklist L);int main(){    int LA[]={3,5,8,11};    int LB[]={2,6,8,9,11,15,20};    LNode L;    int LAlen;    int LBlen;    int i;    LAlen=sizeof(LA)/sizeof(LA[0]);    LBlen=sizeof(LB)/sizeof(LA[0]);    creatList(&L,LA,LAlen);    printList(&L);    for (i=0;i<LBlen;i++)        if(!judgeList(&L,LB[i]))            insertList(&L,LB[i]);    printList(&L);    return 0;}void creatList(Linklist L,int LA[],int len){    /* 用长度为len的数组LA创建链表L */     Linklist p;     Linklist q;     int i;     p=L;     L->next=NULL;     for (i=0;i<len;i++)     {         q=(int *)malloc(sizeof(int));         q->data=LA[i];         q->next=NULL;         p->next=q;         p=p->next;     }     p=NULL;}int judgeList(Linklist L,int x){    /* 判断链表中是否已经存在元素x ，若存在返回1，否则返回0 */    Linklist p;    p=L;    while(p->next)    {        if(p->next->data==x)            return 1;        p=p->next;    }    return 0;}int insertList(Linklist L,int x){    /* 在链表末尾插入元素x */    Linklist p;    Linklist q;    p=L;    while(p->next)        p=p->next;    q=(int *)malloc(sizeof(int));    q->data=x;    q->next=NULL;    p->next=q;}void printList(Linklist L){    /* 输出链表元素 */    Linklist p;    p=L;    while(p->next)    {        printf("%d ",p->next->data);        p=p->next;    }    printf("\n");}3 5 8 113 5 8 11 2 6 9 15 20Press any key to continue