问题描述

c语言数据结构，求算法

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)

解决方案

（C语言-数据结构与算法）还原二叉树
数据结构和算法系列 - c语言归并排序法

解决方案二：
对LA进行遍历，依次把LA中的项加入LB，LC中。依靠修改原LA中项的指针实现。

解决方案三：

 //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
"");    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(""
"");}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
"");    print_link(LB);    printf(""odd
"");    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