数据结构实验之链表五:单链表的拆分
Time Limit: 1000 ms Memory Limit: 65536 KiB
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。
Sample Input
101 3 22 8 15 999 9 44 6 1001
Sample Output
4 622 8 44 6 1 3 15 999 9 1001
Hint
不得使用数组!
拆开,新建两条子链;
#include#include struct node{ int data; struct node *next;};int main(){ struct node *head, *tail, *p; head = (struct node *)malloc(sizeof(struct node)); head->next = NULL; tail = head; int i,n,x=0,y=0; scanf("%d",&n); for(i=0; i data); p->next = NULL; tail->next = p; tail = p; } struct node *head1, *head2, *tail1, *tail2, *p1, *p2; head1 = (struct node *)malloc(sizeof(struct node)); head2 = (struct node *)malloc(sizeof(struct node)); head1->next = NULL; head2->next = NULL; tail1 = head1; tail2 = head2; p = head->next; while(p){ if(!(p->data%2)){ p1 = (struct node *)malloc(sizeof(struct node)); p1->data = p->data; p1->next = NULL; tail1->next = p1; tail1 = p1; x++; } else{ p2 = (struct node *)malloc(sizeof(struct node)); p2->data = p->data; p2->next = NULL; tail2->next = p2; tail2 = p2; y++; } p = p->next; } printf("%d %d\n",x,y); p1 = head1->next; while(p1->next){ printf("%d ",p1->data); p1 = p1->next; } printf("%d\n",p1->data); p2 = head2->next; while(p2->next){ printf("%d ",p2->data); p2 = p2->next; } printf("%d\n",p2->data); return 0;}