PTA作业10单链表6-1 链表拼接

啥都不要说了，做了半天
感天动地

#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 *createlist(){
	struct ListNode *p,*head,*las;
	head=p=(struct ListNode*)malloc(sizeof(struct ListNode));
	while (1){
		scanf("%d",&(p->data));
		if (p->data==-1){
			las->next=NULL;
			return head;
		}
		p->next=(struct ListNode*)malloc(sizeof(struct ListNode));
		las=p;
		p=p->next;
	}
}

struct ListNode *mergelists(struct ListNode *list1, struct ListNode *list2){
	struct ListNode *p,*head,*las;
	if (!list1 && !list2) return NULL;
	head=p=(struct ListNode*)malloc(sizeof(struct ListNode));
	for (;list1;list1=list1->next){
		for (;list2 && list2->data<list1->data;list2=list2->next){
			p->data=list2->data;
			p->next=(struct ListNode*)malloc(sizeof(struct ListNode));
			las=p;
			p=p->next;
		}
		p->data=list1->data;
		p->next=(struct ListNode*)malloc(sizeof(struct ListNode));
		las=p;
		p=p->next;
	}
	for (;list2;list2=list2->next){
		p->data=list2->data;
		p->next=(struct ListNode*)malloc(sizeof(struct ListNode));
		las=p;
		p=p->next;
	}
	las->next=NULL;
	//p=NULL不代表把上一个p的next改NULL，p是个指针，p=NULL只能让p指向NULL，已经创建的空间无法改变 
	return head;
}