数据结构实验之链表四：有序链表的归并

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think:
1顺序建立链表+链表的顺序输出+链表的有序(顺序)归并;链表的元素插入

sdut原题链接

数据结构实验之链表四：有序链表的归并
Time Limit: 1000MS Memory Limit: 65536KB

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。

Example Input
6 5
1 23 26 45 66 99
14 21 28 50 100

Example Output
1 14 21 23 26 28 45 50 66 99 100

Hint
不得使用数组！

Author

以下为accepted代码

#include <stdio.h>#include <string.h>#include <stdlib.h>struct node{    int Data;    struct node *next;};struct node * Build(int n);//顺序建立链表struct node * ans(struct node *head1, struct node *head2);//链表的有序(顺序)归并void Pri(struct node *head);//链表的顺序输出int main(){    int n, m;    while(scanf("%d %d", &n, &m) != EOF)    {        struct node *head1, *head2, *head;        head1 = (struct node *)malloc(sizeof(struct node));        head1->next = NULL;        head2 = (struct node *)malloc(sizeof(struct node));        head2->next = NULL;        head = (struct node *)malloc(sizeof(struct node));        head->next = NULL;        head1 = Build(n);        //Pri(head1);        head2 = Build(m);        //Pri(head2);        head = ans(head1, head2);        Pri(head);    }    return 0;}void Pri(struct node *head)//链表的顺序输出{    struct node *p;    p = head->next;    while(p != NULL)    {        printf("%d%c", p->Data, p->next == NULL? '\n': ' ');        p = p->next;    }}struct node * Build(int n)//顺序建立链表{    struct node *head, *tail;    head = (struct node *)malloc(sizeof(struct node));    head->next = NULL;    tail = head;    struct node *p;    for(int i = 0; i < n; i++)    {        p = (struct node *)malloc(sizeof(struct node));        scanf("%d", &p->Data);        p->next = tail->next;        tail->next = p;        tail = p;    }    return head;}struct node * ans(struct node *head1, struct node *head2)//链表的有序(顺序)归并{    struct node *head, *tail;    head = (struct node *)malloc(sizeof(struct node));    head->next = NULL;    tail = head;    struct node *p, *q, *a, *b;    p = head1->next;    a = head2->next;    q = p, b = a;    while(p != NULL && a != NULL)    {        if(p->Data < a->Data)        {            q = p->next;            p->next = tail->next;            tail->next = p;            tail = p;            p = q;            if(p == NULL)                break;        }        else        {            b = a->next;            a->next = tail->next;            tail->next = a;            tail = a;            a = b;            if(a == NULL)                break;        }    }    if(p == NULL)        tail->next = a;    else if(a == NULL)        tail->next = p;    free(head1);    free(head2);    return head;}/***************************************************User name: Result: AcceptedTake time: 0msTake Memory: 108KBSubmit time: 2017-03-10 17:07:53****************************************************/

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