顺序表应用5：有序顺序表归并
Time Limit: 100 ms Memory Limit: 880 KiB
Submit Statistic
Problem Description
已知顺序表A与B是两个有序的顺序表，其中存放的数据元素皆为普通整型，将A与B表归并为C表，要求C表包含了A、B表里所有元素，并且C表仍然保持有序。
Input
输入分为三行： 第一行输入m、n（1<=m,n<=10000）的值，即为表A、B的元素个数； 第二行输入m个有序的整数，即为表A的每一个元素； 第三行输入n个有序的整数，即为表B的每一个元素；
Output
输出为一行，即将表A、B合并为表C后，依次输出表C所存放的元素。
Sample Input
5 3
1 3 5 6 9
2 4 10
Sample Output
1 2 3 4 5 6 9 10
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
struct node
{
int data;
struct node *next;
};
struct node * Creat_LinkList(int);
struct node * Merge(struct node *, struct node *);
void Output(struct node *);
int main()
{
int m, n;
struct node *head1, *head2;
scanf("%d %d", &m, &n);
head1 = Creat_LinkList(m);
head2 = Creat_LinkList(n);
head1 = Merge(head1, head2);
Output(head1);
return 0;
}
struct node * Creat_LinkList(int n)
{
struct node *head, *p, *tail;
head = (struct node *)malloc(sizeof(struct node));
head->next = NULL;
tail = head;
while(n--)
{
p = (struct node *)malloc(sizeof(struct node));
scanf("%d", &p->data);
p->next = tail->next;
tail->next = p;
tail = p;
}
tail->next = NULL;
return head;
};
struct node *Merge(struct node *head1, struct node *head2)
{
struct node *p1, *p2, *tail;
p1 = head1->next;
p2 = head2->next;
tail = head1;
tail->next = NULL;
free(head2);
while(p1 && p2)
{
if(p1->data < p2->data)
{
tail->next = p1;
p1 = p1->next;
tail = tail->next;
tail->next = NULL;
}
else
{
tail->next = p2;
p2 = p2->next;
tail = tail->next;
tail->next = NULL;
}
}
if(p1 != NULL) tail->next = p1;///此处如果p1不为空，则将p1后面的直接连上去
else tail->next = p2;
return head1;
};
void Output(struct node *head)
{
struct node *p;
p = head->next;
while(p)
{
if(p == head->next) printf("%d", p->data);
else printf(" %d", p->data);
p = p->next;
}
printf("\n");
}
