c语言数据结构，求算法

Question

把一个单链表LA中的奇数项和偶数项分开，分别放在两个单链表LB，LC中(要求利用原空间，头结点空间可另外开辟)

Answer 1

 //输入时以-1结束
#include <stdio.h>
#include <stdlib.h>
struct node
{
    int data;
    struct node * next;
};

struct node * create()
{
    struct node * head = NULL;
    struct node * p = NULL;
    struct node * tmp = NULL;
    int num;
    printf("input number end -1\n");
    while (1)
    {
        scanf("%d", &num);
        if (num == -1)
            break;
        tmp = (struct node *)malloc(sizeof(struct node));
        tmp->data = num;
        if (head == NULL)
        {
            head = tmp;
            p = head;
        }
        else
        {
            p->next = tmp;
            p = p->next;
        }
    }
    p->next = NULL;
    return head;
}

void print_link(struct node * head)
{

    while (head)
    {
        printf("%d ", head->data);
        head = head->next;
    }
    printf("\n");
}

struct node ** devide_link(struct node * head)
{
    struct node *head_even_tmp = NULL;
    struct node *head_odd_tmp = NULL;
    struct node ** head_result = (struct node * *)malloc(2 * sizeof(struct node *));
    head_result[0] = NULL;
    head_result[1] = NULL;
    while (head)
    {
        if (head->data % 2) //偶数
        {
            if (head_result[0] == NULL)
            {
                head_result[0] = head;
                head_even_tmp = head;
            }
            else {
                head_even_tmp->next = head;
                head_even_tmp = head_even_tmp->next;
            }
        }
        else  //奇数
        {
            if (head_result[1] == NULL)
            {
                head_result[1] = head;
                head_odd_tmp = head;
            }
            else {
                head_odd_tmp->next = head;
                head_odd_tmp = head_odd_tmp->next;
            }
        }
        head = head->next;
    }
    if(head_odd_tmp)
        head_odd_tmp->next = NULL;
    if(head_even_tmp)
        head_even_tmp->next = NULL;
    return head_result;
}

int main()
{
    struct node * LA = create();
    struct node ** head_result = NULL; 
    struct node *LB, *LC;

    print_link(LA);
    head_result = devide_link(LA);
    LB = head_result[0];
    LC = head_result[1];
    printf("even\n");
    print_link(LB);
    printf("odd\n");
    print_link(LC);
}
追问：
不需要整个程序，只要写一段实现要求的算法代码
追答：
那你只需要给出链表节点定义和 分离函数即可
1.链表节点定义
struct node
{
int data;
struct node * next;
};
1.将链表分为奇数，偶数两个链表
struct node ** devide_link(struct node * head)
{
struct node *head_even_tmp = NULL;
struct node *head_odd_tmp = NULL;
struct node ** head_result = (struct node * *)malloc(2 * sizeof(struct node *));
head_result[0] = NULL;
head_result[1] = NULL;
while (head)
{
if (head->data % 2) //偶数
{
if (head_result[0] == NULL)
{
head_result[0] = head;
head_even_tmp = head;
}
else {
head_even_tmp->next = head;
head_even_tmp = head_even_tmp->next;
}
}
else //奇数
{
if (head_result[1] == NULL)
{
head_result[1] = head;
head_odd_tmp = head;
}
else {
head_odd_tmp->next = head;
head_odd_tmp = head_odd_tmp->next;
}
}
head = head->next;
}
if(head_odd_tmp)
head_odd_tmp->next = NULL;
if(head_even_tmp)
head_even_tmp->next = NULL;
return head_result;
}
追问：
好的，谢谢啦