【c语言篇】每日一题-pta-实验11-2-5 链表拼接

题目如下：

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 */

 

 代码如下：

struct ListNode *mergelists(struct ListNode *list1, struct ListNode *list2)
{
    //创建一个新的头结点作为新链表的结点
    struct ListNode *newhead = (struct ListNode*)malloc(sizeof(struct ListNode));
    struct ListNode *current = newhead;
    while(list1 != NULL && list2 !=NULL)
    {
        if(list1->data < list2->data)
        {
            current->next = list1;
            list1 = list1->next;
        }
        else{
            current->next = list2;
            list2 = list2->next;
        }
        current = current->next;
    }
    //将l1或l2的剩余结点连接到新链表尾部
    current->next = list1 == NULL ?list2:list1;
    //头结点不存储数据，返回头结点的下一个结点
    return newhead->next;
}