将两个非递减的有序链表合成为一个非递增的有序链表 c语言

可以按照以下步骤来实现：

1. 定义两个指针，分别指向两个链表的头节点，以及一个指针，指向合并后链表的头节点。
2. 比较两个链表当前节点的值，将较小的节点加入到合并后的链表中，并将指针指向下一个节点。
3. 循环执行步骤2，直到有一个链表遍历完毕。
4. 将剩余的链表连接到合并后的链表的末尾。
5. 返回合并后的链表的头节点。

以下是示例代码实现：

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

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

Node *mergeLists(Node *list1, Node *list2) {
    Node *mergedList = NULL;
    Node *currentNode = NULL;
    
    while (list1 != NULL && list2 != NULL) {
        Node *minNode;
        if (list1->val <= list2->val) {
            minNode = list1;
            list1 = list1->next;
        } else {
            minNode = list2;
            list2 = list2->next;
        }
        
        if (mergedList == NULL) {
            mergedList = currentNode = minNode;
        } else {
            currentNode->next = minNode;
            currentNode = minNode;
        }
    }
    
    if (list1 != NULL) {
        if (mergedList == NULL) {
            mergedList = list1;
        } else {
            currentNode->next = list1;
        }
    }
    
    if (list2 != NULL) {
        if (mergedList == NULL) {
            mergedList = list2;
        } else {
            currentNode->next = list2;
        }
    }
    
    // 反转链表
    Node *prev = NULL;
    Node *curr = mergedList;
    Node *nextNode;
    while (curr != NULL) {
        nextNode = curr->next;
        curr->next = prev;
        prev = curr;
        curr = nextNode;
    }
    mergedList = prev;
    
    return mergedList;
}

int main() {
    Node *list1 = (Node *) malloc(sizeof(Node));
    Node *list2 = (Node *) malloc(sizeof(Node));
    
    list1->val = 1;
    Node *node2 = (Node *) malloc(sizeof(Node));
    node2->val = 3;
    Node *node3 = (Node *) malloc(sizeof(Node));
    node3->val = 5;
    list1->next = node2;
    node2->next = node3;
    node3->next = NULL;
    
    list2->val = 2;
    Node *node5 = (Node *) malloc(sizeof(Node));
    node5->val = 4;
    Node *node6 = (Node *) malloc(sizeof(Node));
    node6->val = 6;
    list2->next = node5;
    node5->next = node6;
    node6->next = NULL;
    
    Node *mergedList = mergeLists(list1, list2);
    
    Node *currentNode = mergedList;
    while (currentNode != NULL) {
        printf("%d ", currentNode->val);
        currentNode = currentNode->next;
    }
    printf("\n");
    
    return 0;
}

注：该代码实现中，为了将合并后的链表变成非递增的有序链表，使用了链表反转的方法，将链表从头到尾遍历一遍，将节点的next指针指向前一个节点，最后将链表的头节点指向最后一个节点。