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;
}