数据结构实验之链表四:有序链表的归并
Time Limit: 1000 ms Memory Limit: 65536 KiB
Problem Description
分别输入两个有序的整数序列(分别包含M和N个数据),建立两个有序的单链表,将这两个有序单链表合并成为一个大的有序单链表,并依次输出合并后的单链表数据。
Input
第一行输入M与N的值;
第二行依次输入M个有序的整数;
第三行依次输入N个有序的整数。
Output
输出合并后的单链表所包含的M+N个有序的整数。
Sample Input
6 5 1 23 26 45 66 99 14 21 28 50 100
Sample Output
1 14 21 23 26 28 45 50 66 99 100
Hint
不得使用数组!
Source
#include <stdio.h>
#include <stdlib.h>
struct node
{
int data;
struct node *next;
};
int main()
{
struct node *head, *head1, *head2, *tail1, *tail2, *tail, *p;
int m, n, i;
scanf("%d%d", &m, &n);
head1 = (struct node *)malloc(sizeof(struct node));
head1->next = NULL;
tail = head1;
for(i = 1; i <= m; i++)
{
p = (struct node *)malloc(sizeof(struct node));
scanf("%d", &p->data);
p->next = NULL;
tail->next = p;
tail = p;
}
head2 = (struct node *)malloc(sizeof(struct node));
head2->next = NULL;
tail = head2;
for(i = 1; i <= n; i++)
{
p = (struct node *)malloc(sizeof(struct node));
scanf("%d", &p->data);
p->next = NULL;
tail->next = p;
tail = p;
}
head = (struct node *)malloc(sizeof(struct node));
head->next = NULL;
tail = head;
tail1 = head1->next;
tail2 = head2->next;
for(i = 1; i <= m + n; i++)
{
if(tail1 == NULL)
{
tail->next = tail2;
tail = tail2;
tail2 = tail2->next;
}
else if(tail2 == NULL)
{
tail->next = tail1;
tail = tail1;
tail1 = tail1->next;
}
else
{
if(tail1->data < tail2->data)
{
tail->next = tail1;
tail = tail1;
tail1 = tail1->next;
}
else
{
tail->next = tail2;
tail = tail2;
tail2 = tail2->next;
}
}
}
tail->next = NULL;
for(tail = head->next; tail->next != NULL; tail = tail->next)
{
printf("%d ", tail->data);
}
printf("%d\n", tail->data);
return 0;
}