数据结构实验之链表四:有序链表的归并
2018-03-24 16:57
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数据结构实验之链表四:有序链表的归并
Time Limit: 1000 msMemory Limit: 65536 KiB
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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
#include<stdio.h>
#include<stdlib.h>
struct node
{
int data;
struct node *next;
}*head1, *head2;
struct node *creat1(int m)
{
int i;
struct node *p, *tail1;
head1 = (struct node*)malloc(sizeof(struct node));
head1-> next = NULL;
tail1 = head1;
for(i = 1; i <= m; i++)
{
p = (struct node*)malloc(sizeof(struct node));
p-> next = NULL;
scanf("%d", &p-> data);
tail1-> next = p;
tail1 = p;
}
return head1;
}
struct node *creat2(int n)
{
int i;
struct node *q, *tail2;
head2 = (struct node*)malloc(sizeof(struct node));
head2-> next = NULL;
tail2 = head2;
for(i = 1; i <= n; i++)
{
q = (struct node*)malloc(sizeof(struct node));
q-> next = NULL;
scanf("%d", &q-> data);
tail2-> next = q;
tail2 = q;
}
return head2;
}
struct node *merge(struct node *head1, struct node *head2)
{
struct node *q, *p, *tail;
p = head1-> next;
q = head2-> next;
head1-> next = NULL;
tail = head1;
free(head2);
while(p && q)
{
if(p-> data > q-> data)
{
tail-> next = q;
tail = q;
q = q-> next;
tail-> next = NULL;
}
else
{
tail-> next = p;
tail = p;
p = p-> next;
tail-> next = NULL;
}
}
if(p)
{
tail-> next = p;
}
else
{
tail-> next = q;
}
return head1;
}
void print(struct node *head1)
{
struct node *p;
p = head1-> next;
while(p != NULL)
{
if(p-> next != NULL)
{
printf("%d ", p-> data);
}
else
{
printf("%d\n", p-> data);
}
p = p-> next;
}
}
int main(void)
{
int m, n;
struct node *h1, *h2, *h3;
scanf("%d %d", &m, &n);
h1 = creat1(m);
h2 = creat2(n);
h3 = merge(h1, h2);
print(h3);
return 0;
}
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