int flag[10000+10];
void quchong(LNode *L){LNode *p = L;while (p->next!=NULL){int t = p->next->data;if (t<0) t = -t;if (flag[t]==0){flag[t] = 1;p = p->next;}else{LNode *temp = p->next;p->next = temp->next;free(temp);}}
}

//合并之后,放到第一个链表里面
void _merge(LNode *h1,LNode *h2){LNode *p1 = h1->next;LNode *p2 = h2->next,*t;h1->next = NULL;t = h1;while (p1!=NULL && p2!=NULL){if (p1->data<p2->data){t->next = p1;t = p1;p1 = p1->next;}else{t->next = p2;t = p2;p2 = p2->next;}}while (p1!=NULL) {t->next = p1;t = p1;p1=p1->next;}while (p2!=NULL){t->next = p2;t = p2;p2 = p2->next;}
}

//先把第二个链表原地倒置 然后当成2.6的合并做就好
void _merge(LNode *h1,LNode *h2){LNode *p = h2->next;h2->next = NULL;while (p){LNode * temp = p->next;p->next = h2->next;h2->next = p;p = temp;}LNode *p1 = h1->next;LNode *p2 = h2->next,*t;h1->next = NULL;t = h1;while (p1!=NULL && p2!=NULL){if (p1->data<p2->data){t->next = p1;t = p1;p1 = p1->next;}else{t->next = p2;t = p2;p2 = p2->next;}}while (p1!=NULL) {t->next = p1;t = p1;p1=p1->next;}while (p2!=NULL){t->next = p2;t = p2;p2 = p2->next;}
}

//这里必须要用二重指针,因为我们要改变第一个链表的指向(我们新创的一个链表)
//main函数里调用_merge(&h1,h2)即可(h2不用变)
void _merge(LNode **h1,LNode *h2){LNode *p1 = *h1;LNode *p2 = h2,*h3;h3 = newnode();//不好直接接在第一个链表上,所以新创建一个LNode *t = newnode();t = h3;while (p1!=NULL && p2!=NULL){if (p1->data<p2->data){LNode *temp = newnode(); //不能破坏原始链表,所以只能自己创了temp->data = p1->data;t->next = temp;t = temp;p1 = p1->next;}else if (p1->data>p2->data){LNode *temp = newnode();temp->data = p2->data;t->next = temp;t = temp;p2 = p2->next;}else if (p1->data==p2->data){//如果两个链表中有,那么只要第一个链表中的p2 = p2->next;}}if (p1==NULL) p1 = p2; //这样写比较简洁while (p1!=NULL) {LNode *temp = newnode();temp->data = p1->data;t->next = temp;t = temp;p1=p1->next;}*h1 = h3->next;//让h1再指向h3
}

void _merge(LNode *h1,LNode *h2){LNode *p1 = h1->next;LNode *p2 = h2->next,*t;h1->next = NULL;t = h1;while (p1!=NULL && p2!=NULL){if (p1->data<p2->data){p1 = p1->next;}else if (p1->data>p2->data){p2 = p2->next;}else if (p1->data==p2->data){//如果两个链表中有,任取一个放在newA链表后面t->next = p1;t = p1;p1 = p1->next;}}
}