将两个无序的链表合并在一起，并去除其中的重复元素，希望时间复杂度尽可能小，应该使用什么算法？请给出C语言示例

可以使用哈希表来实现。具体做法是，遍历两个链表，将每个元素的值作为键值存入哈希表中，如果发现重复元素，则将其从链表中删除。最后将两个链表合并即可。

以下是C语言示例代码：

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

#define HASH_SIZE 10000

typedef struct ListNode {
    int val;
    struct ListNode* next;
} ListNode;

int hash(int val) {
    return abs(val) % HASH_SIZE;
}

ListNode* mergeLists(ListNode* l1, ListNode* l2) {
    ListNode* head = NULL;
    ListNode* tail = NULL;
    ListNode* p = l1 ? l1 : l2;
    ListNode* q = l1 && l2 ? l2 : NULL;
    int h[HASH_SIZE] = {0};

    while (p) {
        int key = hash(p->val);
        if (h[key] == 0) {
            h[key] = 1;
            if (!head) {
                head = tail = p;
            } else {
                tail->next = p;
                tail = p;
            }
        }
        p = p->next;
    }

    while (q) {
        int key = hash(q->val);
        if (h[key] == 0) {
            h[key] = 1;
            if (!head) {
                head = tail = q;
            } else {
                tail->next = q;
                tail = q;
            }
        }
        q = q->next;
    }

    if (tail) {
        tail->next = NULL;
    }

    return head;
}

void printList(ListNode* head) {
    ListNode* p = head;
    while (p) {
        printf("%d ", p->val);
        p = p->next;
    }
    printf("\n");
}

int main() {
    ListNode* l1 = malloc(sizeof(ListNode));
    l1->val = 1;
    l1->next = malloc(sizeof(ListNode));
    l1->next->val = 2;
    l1->next->next = malloc(sizeof(ListNode));
    l1->next->next->val = 3;
    l1->next->next->next = NULL;

    ListNode* l2 = malloc(sizeof(ListNode));
    l2->val = 2;
    l2->next = malloc(sizeof(ListNode));
    l2->next->val = 3;
    l2->next->next = malloc(sizeof(ListNode));
    l2->next->next->val = 4;
    l2->next->next->next = NULL;

    ListNode* head = mergeLists(l1, l2);
    printList(head);

    return 0;
}