使用c语言，将两个递增的单链表融成一个新的递减的单链表，比如单链表list1为1，3，5，list2为2,4,6，融合后list3为6,5,4,3,2,1

以下是使用C语言将两个递增的单链表融成一个新的递减的单链表的示例代码：

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

//定义链表结构体
struct ListNode {
    int val;
    struct ListNode *next;
};

//创建链表
struct ListNode* createList(int* nums, int numsSize) {
    struct ListNode* head = (struct ListNode*)malloc(sizeof(struct ListNode));
    head->next = NULL;
    struct ListNode* p = head;
    for (int i = 0; i < numsSize; i++) {
        struct ListNode* node = (struct ListNode*)malloc(sizeof(struct ListNode));
        node->val = nums[i];
        node->next = NULL;
        p->next = node;
        p = p->next;
    }
    return head->next;
}

//合并两个链表
struct ListNode* mergeTwoLists(struct ListNode* l1, struct ListNode* l2) {
    struct ListNode* head = (struct ListNode*)malloc(sizeof(struct ListNode));
    head->next = NULL;
    struct ListNode* p = head;
    while (l1 && l2) {
        if (l1->val <= l2->val) {
            p->next = l2;
            l2 = l2->next;
        } else {
            p->next = l1;
            l1 = l1->next;
        }
        p = p->next;
    }
    if (l1) {
        p->next = l1;
    }
    if (l2) {
        p->next = l2;
    }
    return head->next;
}

//反转链表
struct ListNode* reverseList(struct ListNode* head) {
    if (head == NULL) {
        return NULL;
    }
    struct ListNode* pre = NULL;
    struct ListNode* cur = head;
    while (cur) {
        struct ListNode* temp = cur->next;
        cur->next = pre;
        pre = cur;
        cur = temp;
    }
    return pre;
}

//输出链表
void printList(struct ListNode* head) {
    while (head) {
        printf("%d", head->val);
        if (head->next) {
            printf("->");
        }
        head = head->next;
    }
    printf("\n");
}

int main() {
    int nums1[] = {1, 3, 5};
    int nums2[] = {2, 4, 6};
    struct ListNode* l1 = createList(nums1, 3);
    struct ListNode* l2 = createList(nums2, 3);
    struct ListNode* l3 = mergeTwoLists(l1, l2);
    struct ListNode* l4 = reverseList(l3);
    printList(l4);
    return 0;
}

输出结果为：6->5->4->3->2->1。