有两个链表，将这两个链表排序后组成新的链表

版权声明：所有博客均由作者本人原创，若要转载，请注明出处。谢谢 https://blog.csdn.net/LU_Leo/article/details/83215398

如 list1={2,4,5,6,8,9};list2={1,3,7};

合并的新链表list3={1,2,3,4,5,6,7,8,9}

面试的时候脑子有点乱，没有写出来，答案可能不完善，有漏洞，如有发现，请留言。

#include <iostream>
#include <stdio.h>
using namespace std;
struct ListNode//链表定义
{
	int val;
	struct ListNode *next;
};
ListNode* create(int arr[], int n)//丛数组读取元素组成链表
{
	if (arr == NULL || n <= 0)
		return NULL;

	ListNode *head = (ListNode*)malloc(sizeof(struct ListNode));
	head->val = arr[0];
	head->next = NULL;
	ListNode *tmp = head;
	ListNode *next = NULL;
	for (int i = 1; i < n; i++)
	{
		next = (ListNode*)malloc(sizeof(struct ListNode));
		if (next)
		{
			next->val = arr[i];
			next->next = NULL;
			tmp->next = next;
			tmp = next;
		}
	}
	return head;
}
//调用链表打印函数，打印出合并后的链表元素，代码如下：
void printList(struct ListNode *head)
{
	if (head == NULL)
		return;
	while (head)
	{
		printf("%d ", head->val);
		head = head->next;
	}
}
ListNode* createList(struct ListNode * list1, struct ListNode* list2)//合并 并排好序
{
	if (list1 == NULL && list2 == NULL)
		return NULL;
	ListNode *head = (ListNode*)malloc(sizeof(struct ListNode));//记录链表头指针
	if (list1->val < list2->val)
	{
		head->val = list1->val;
		list1 = list1->next;
		head->next = NULL;
	}
	else
	{
		head->val = list2->val;
		list2 = list2->next;
		head->next = NULL;
	}
	ListNode *temp = head;//记录当前链表指针位置
	ListNode *next = NULL;//记录正在操作的元素指针
	while (list1&&list2) {
		next = (ListNode*)malloc(sizeof(struct ListNode));
		if (list1->val < list2->val)
		{
			next->val = list1->val;
			next->next = NULL;
			list1 = list1->next;
			temp->next = next;
			temp = next;
		}
		else
		{
			next->val = list2->val;
			next->next = NULL;
			list2 = list2->next;
			temp->next = next;
			temp = next;
		}
	}
	while (list1)
	{
		next = (ListNode*)malloc(sizeof(struct ListNode));
	    next->val = list1->val;
		next->next = NULL;
		list1 = list1->next;
		temp->next = next;
		temp = next;
	}
	while (list2)
	{
		next = (ListNode*)malloc(sizeof(struct ListNode));
		next->val = list2->val;
		next->next = NULL;
		list2 = list2->next;
		temp->next = next;
		temp = next;
	}
	return head;//返回链表头指针
}
int main()
{
	int arr1[] = { 2,4,5,6,8,9 };
	int arr2[] = { 1,3,7 };
	ListNode* list1 = create(arr1, 6);
	ListNode* list2 = create(arr2, 3);
	printList(list1);
	cout << endl;
	printList(list2);
	ListNode* list3= createList(list1,list2);
	cout << endl;
	printList(list3);
}