实验11-2-6 奇数值结点链表 (20分)
本题要求实现两个函数,分别将读入的数据存储为单链表、将链表中奇数值的结点重新组成一个新的链表。链表结点定义如下:
struct ListNode {
int data;
ListNode *next;
};
函数接口定义:
struct ListNode *readlist();
struct ListNode *getodd( struct ListNode **L );
函数readlist从标准输入读入一系列正整数,按照读入顺序建立单链表。当读到−1时表示输入结束,函数应返回指向单链表头结点的指针。
函数getodd将单链表L中奇数值的结点分离出来,重新组成一个新的链表。返回指向新链表头结点的指针,同时将L中存储的地址改为删除了奇数值结点后的链表的头结点地址(所以要传入L的指针)。
裁判测试程序样例:
#include <stdio.h>
#include <stdlib.h>
struct ListNode {
int data;
struct ListNode *next;
};
struct ListNode *readlist();
struct ListNode *getodd( struct ListNode **L );
void printlist( struct ListNode *L )
{
struct ListNode *p = L;
while (p) {
printf("%d ", p->data);
p = p->next;
}
printf("\n");
}
int main()
{
struct ListNode *L, *Odd;
L = readlist();
Odd = getodd(&L);
printlist(Odd);
printlist(L);
return 0;
}
输入样例:
1 2 2 3 4 5 6 7 -1
输出样例:
1 3 5 7
2 2 4 6
struct ListNode *readlist()
{
struct ListNode *p, *head = NULL, *tail;
int a;
scanf("%d", &a);
while(a != -1)
{
p = (struct ListNode*)malloc(sizeof(struct ListNode));
p -> data = a;
p -> next = NULL;
if(head == NULL) head = p;
else tail -> next = p;
tail = p;
scanf("%d", &a);
}
return head;
}
struct ListNode *getodd( struct ListNode **L )
{
struct ListNode *p, *p1, *p2, *h1 = NULL, *h2 = NULL;
int a;
while(*L)
{
a = (*L) -> data;
if( a % 2 == 1 )
{
p = (struct ListNode*)malloc(sizeof(struct ListNode));
p -> data = a;
p -> next = NULL;
if(h1 == NULL) h1 = p;
else p1 -> next = p;
p1 = p;
}
else
{
p=(struct ListNode*)malloc(sizeof(struct ListNode));
p -> data = a;
p -> next = NULL;
if(h2 == NULL) h2 = p;
else p2 -> next = p;
p2 = p;
}
*L = (*L)->next;
}
*L = h2;
return h1;
}