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;
}