6-3 链表拼接 (20分)

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本文介绍了一个简单的函数，用于合并两个按升序排列的链表。通过实现mergelists函数，可以将两个有序链表合并为一个新的有序链表。示例展示了如何使用此函数将两个链表合并，并输出合并后的链表。

6-3 链表拼接 (20分)
本题要求实现一个合并两个有序链表的简单函数。链表结点定义如下：

struct ListNode {
    int data;
    struct ListNode *next;
};

函数接口定义：

struct ListNode *mergelists(struct ListNode *list1, struct ListNode *list2);

其中list1和list2是用户传入的两个按data升序链接的链表的头指针；函数mergelists将两个链表合并成一个按data升序链接的链表，并返回结果链表的头指针。

裁判测试程序样例：

#include <stdio.h>
#include <stdlib.h>

struct ListNode {
    int data;
    struct ListNode *next;
};

struct ListNode *createlist(); /*裁判实现，细节不表*/
struct ListNode *mergelists(struct ListNode *list1, struct ListNode *list2);
void printlist( struct ListNode *head )
{
     struct ListNode *p = head;
     while (p) {
           printf("%d ", p->data);
           p = p->next;
     }
     printf("\n");
}

int main()
{
    struct ListNode  *list1, *list2;

    list1 = createlist();
    list2 = createlist();
    list1 = mergelists(list1, list2);
    printlist(list1);
	
    return 0;
}

/* 你的代码将被嵌在这里 */

输入样例：
1 3 5 7 -1
2 4 6 -1
输出样例：
1 2 3 4 5 6 7

#include <stdio.h>
#include <stdlib.h>

struct ListNode {
    int data;
    struct ListNode *next;

};

struct ListNode *createlist(); /*裁判实现，细节不表*/
struct ListNode *mergelists(struct ListNode *list1, struct ListNode *list2);

void printlist(struct ListNode *head) {
    struct ListNode *p = head;
    while (p) {
        printf("%d ", p->data);
        p = p->next;
    }
    printf("\n");
}

int main() {
    struct ListNode *list1, *list2;

    list1 = createlist();
    list2 = createlist();
    list1 = mergelists(list1, list2);
    printlist(list1);

    return 0;
}
struct ListNode *newNode(int x) ;
struct ListNode *listInsrt(struct ListNode *list, int data, struct ListNode **rear) {
    if (list == NULL) {
        list = newNode(data);
        *rear = list;
        return list;
    }
    (*rear)->next = newNode(data);
    *rear = (*rear)->next;
    return list;

}

struct ListNode *createlist() {
    struct ListNode *list = NULL, *rear = NULL;
    int x;
    while (scanf("%d", &x) && x != -1) {
        list = listInsrt(list, x, &rear);
    }
    return list;

}

/* 你的代码将被嵌在这里 */

struct ListNode *newNode(int x) {
    struct ListNode *a = (struct ListNode *) malloc(sizeof(struct ListNode));
    a->data = x;
    a->next = NULL;
    return a;
}

struct ListNode *little(struct ListNode *newlist, int data, struct ListNode **rear) {

    struct ListNode *newN = newNode(data);
    if (newlist == NULL)newlist = newN;
    else {
        if (newlist->next == NULL) {
            newlist->next = newN;
            *rear = newlist->next;
            newlist->next = *rear;
        } else {
            (*rear)->next = newN;
            *rear = (*rear)->next;
        }
    }
    return newlist;

}

struct ListNode *mergelists(struct ListNode *list1, struct ListNode *list2) {
    struct ListNode *newlist = NULL, *rear = newlist;
    while (list1 && list2) {
        while (list1 && list2 && list1->data < list2->data) {
            newlist = little(newlist, list1->data, &rear);
            list1 = list1->next;
        }
        while (list1 && list2 && list1->data >= list2->data) {
            newlist = little(newlist, list2->data, &rear);
            list2 = list2->next;
        }
    }
    while (list1) {
        newlist = little(newlist, list1->data, &rear);
        list1 = list1->next;
    }
    while (list2) {
        newlist = little(newlist, list2->data, &rear);
        list2 = list2->next;
    }
    return newlist;
}