利用c语言将两个循环单链表合并为一个循环单链表，且空间复杂度为O（1），写出完整代码

首先需要定义循环单链表的结构体和节点结构体：

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

typedef struct LinkedList {
    Node* head;
} LinkedList;

然后定义一个函数来将两个循环单链表合并为一个循环单链表：

LinkedList* mergeLinkedList(LinkedList* list1, LinkedList* list2) {
    if (list1->head == NULL) {
        return list2;
    }
    if (list2->head == NULL) {
        return list1;
    }
    Node* list1Tail = list1->head;
    while (list1Tail->next != list1->head) {
        list1Tail = list1Tail->next;
    }
    Node* list2Tail = list2->head;
    while (list2Tail->next != list2->head) {
        list2Tail = list2Tail->next;
    }
    list1Tail->next = list2->head;
    list2Tail->next = list1->head;
    free(list2);
    return list1;
}

该函数首先判断两个链表是否为空，如果其中一个为空，则返回另一个链表。接着，找到两个链表的尾节点，分别为list1Tail和list2Tail。然后将list2的头节点插入list1的尾部，再将list1的头节点插入list2的尾部，即可完成合并。最后释放list2的内存空间，返回list1。

完整代码如下：

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

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

typedef struct LinkedList {
    Node* head;
} LinkedList;

LinkedList* mergeLinkedList(LinkedList* list1, LinkedList* list2) {
    if (list1->head == NULL) {
        return list2;
    }
    if (list2->head == NULL) {
        return list1;
    }
    Node* list1Tail = list1->head;
    while (list1Tail->next != list1->head) {
        list1Tail = list1Tail->next;
    }
    Node* list2Tail = list2->head;
    while (list2Tail->next != list2->head) {
        list2Tail = list2Tail->next;
    }
    list1Tail->next = list2->head;
    list2Tail->next = list1->head;
    free(list2);
    return list1;
}

int main() {
    LinkedList* list1 = (LinkedList*)malloc(sizeof(LinkedList));
    LinkedList* list2 = (LinkedList*)malloc(sizeof(LinkedList));
    LinkedList* mergedList;

    Node* node1 = (Node*)malloc(sizeof(Node));
    node1->data = 1;
    Node* node2 = (Node*)malloc(sizeof(Node));
    node2->data = 2;
    Node* node3 = (Node*)malloc(sizeof(Node));
    node3->data = 3;

    node1->next = node2;
    node2->next = node3;
    node3->next = node1;
    list1->head = node1;

    Node* node4 = (Node*)malloc(sizeof(Node));
    node4->data = 4;
    Node* node5 = (Node*)malloc(sizeof(Node));
    node5->data = 5;
    Node* node6 = (Node*)malloc(sizeof(Node));
    node6->data = 6;

    node4->next = node5;
    node5->next = node6;
    node6->next = node4;
    list2->head = node4;

    mergedList = mergeLinkedList(list1, list2);

    Node* currentNode = mergedList->head;
    do {
        printf("%d ", currentNode->data);
        currentNode = currentNode->next;
    } while (currentNode != mergedList->head);
    printf("\n");

    free(node1);
    free(node2);
    free(node3);
    free(node4);
    free(node5);
    free(node6);
    free(list1);
    free(list2);
    free(mergedList);
    return 0;
}