SDUT 2120-数据结构实验之链表五：单链表的拆分

Problem Description

输入N个整数顺序建立一个单链表，将该单链表拆分成两个子链表，第一个子链表存放了所有的偶数，第二个子链表存放了所有的奇数。两个子链表中数据的相对次序与原链表一致。

Input

第一行输入整数N;；
第二行依次输入N个整数。

Output

第一行分别输出偶数链表与奇数链表的元素个数；
第二行依次输出偶数子链表的所有数据；
第三行依次输出奇数子链表的所有数据。

Example Input

10
1 3 22 8 15 999 9 44 6 1001

Example Output

4 6
22 8 44 6
1 3 15 999 9 1001

#include<stdio.h>#include<stdlib.h>typedef struct node{    int data;    struct node *next;}ST;int n1, n2;ST *creat(int n){    ST *tail, *p, *head;    head = (ST *)malloc(sizeof(ST));    head->next = NULL;    tail = head;    while(n--)    {        p = (ST *)malloc(sizeof(ST));        scanf("%d", &p->data);        p->next = NULL;        tail->next = p;        tail = p;    }    return head;}void Break(ST *head, ST *head1, ST *head2){    ST *tail, *p, *q, *tai;    p = head1;    q = head2;    tail = head->next;    free(head);    tai = tail->next;    for(; tail != NULL; )    {        tail->next = NULL;        if(tail->data % 2 == 0)        {            p->next = tail;            p = p->next;            n1++;        }        else        {            q->next = tail;            q = q->next;            n2++;        }        tail = tai;        if(tail) tai = tai->next;    }}void input(ST *head){    ST *tail;    for(tail = head->next; tail != NULL; tail = tail->next)    {        printf("%d", tail->data);        if(tail->next != NULL) printf(" ");        else printf("\n");    }}int main(){    ST *head, *head1, *head2;    int n;    scanf("%d", &n);    head = creat(n);    head1 = (ST *)malloc(sizeof(ST)); head1->next = NULL;    head2 = (ST *)malloc(sizeof(ST)); head2->next = NULL;    n1 = n2 = 0;    Break(head, head1, head2);    printf("%d %d\n", n1, n2);    input(head1);    input(head2);    return 0;}