C. DS单链表--合并

【id:23】【20分】C. DS单链表–合并
题目描述

假定两个单链表是递增有序，定义并实现以下函数，完成两个单链表的合并，继续保持递增有序

int LL_merge(ListNode *La, ListNode *Lb)

输入

第1行先输入n表示有n个数据，接着输入n个数据
第2行先输入m表示有M个数据，接着输入m个数据

输出

输出合并后的单链表数据，数据之间用空格隔开

样例查看模式
正常显示
查看格式
输入样例1 <-复制
3 11 33 55n
4 22 44 66 88n
输出样例1
11 22 33 44 55 66 88 n

#include
#include
#include
#include
using namespace std;

class ListNode {
public:
	int data;
	ListNode* next;
	ListNode()
	{
		next = NULL;
	}
	ListNode(int i){
		this->data=i;
		next=NULL;
	} 
};

class LinkList {
public:
	ListNode* head;
	int len;
	//操作定义
	LinkList();
	~LinkList();
	LinkList* LL_index(int i);//返回第i个节点的指针，如果不存在返回NULL
	int LL_get(int i); //获取第i个元素的数据
	int LL_insrt(int i, int item);
	int LL_del(int i);
	void LL_display();
	int LL_swap(int  pa, int pb){
		if(pa>pb) return LL_swap(pb,pa);
		if (pa<1 || pa>len||pb<1||pb>len) return 0;
		ListNode* cur1, * pre1=head, *cur2,*pre2=pre1,*tail;
		cur1 = head->next;
		cur2 = head->next;
		for (int q = 1; q < pa; q++)
		{
			if (cur1 != NULL)
			{
				pre1 = cur1;
				cur1 = cur1->next;
			}
		}
		for (int q = 1; q < pb; q++)
		{
			if (cur2 != NULL)
			{
				pre2 = cur2;
				cur2 = cur2->next;
			}
		}
		tail=cur2->next; 
		cur2->next=cur1->next;
		pre1->next=cur2;
		cur1->next=tail;
		pre2->next=cur1;
		return 1;
	}
	
};

LinkList::LinkList()
{
	head = new ListNode();
	len = 0;
}

LinkList::~LinkList()
{
	ListNode* p, * q;
	p = head;
	while (p != NULL)
	{
		q = p;
		p = p->next;
		delete q;
	}
	len = 0;
	head = NULL;
}

void LinkList::LL_display()
{
	ListNode* p;
	p = head->next;
	while (p)
	{
		cout << p->data << " ";
		p = p->next;
	}
	cout << endl;
}

int LinkList::LL_insrt(int i, int item)
{
	if (i<1 || i>len + 1) return 0;
	ListNode* cur, * pre=head, * now;
	cur = head->next;
	for (int q = 1; q < i; q++)
	{
		if (cur != NULL)
		{
			pre = cur;
			cur = cur->next;
		}
	}
	now = new ListNode;
	now->data = item;
	pre->next = now;
	now->next = cur;
	len++;
	return 1;
}

int LinkList::LL_del(int i)
{
	if (i<1 || i>len) return 0;
	ListNode* cur, * pre = head, * now;
	cur = head->next;
	for (int q = 1; q < i; q++)
	{
		if (cur != NULL)
		{
			pre = cur;
			cur = cur->next;
		}
	}
	pre->next = cur->next;
	len--;
	//delete cur;
	return 1;
}

int LinkList::LL_get(int i)
{
	if (i<1 || i>len) return 0;
	ListNode* cur, * pre = head, * now;
	cur = head->next;
	for (int q = 1; q < i; q++)
	{
		if (cur != NULL)
		{
			pre = cur;
			cur = cur->next;
		}
	}
	return cur->data;
}

ListNode* LL_merge(ListNode *l1, ListNode *l2){
	ListNode* preHead = new ListNode(-1);
	ListNode* prev = preHead;
    while (l1 != NULL && l2 != NULL) {
        if (l1->data < l2->data) {
                prev->next = l1;
                //cout<data<<" ";
                l1 = l1->next;
            } else {
                prev->next = l2;
                //cout<data<<" ";
				l2 = l2->next;
            }
            prev = prev->next;
        }
        // 合并后 l1 和 l2 最多只有一个还未被合并完，我们直接将链表末尾指向未合并完的链表即可
        prev->next = l1 == NULL ? l2 : l1;
        return preHead->next;
}
int main()
{
	ListNode *l1,*l2,*l3,*cur;
	l1=new ListNode(-1);
	l2=new ListNode(-1);
	int num,lenth;
	cin >> lenth;
	cur=l1;
	for (int i = 0; i < lenth; i++)
	{
		cin >> num;
		cur->next=new ListNode(num);
		cur=cur->next;
	}
	cin >> lenth;
	cur=l2;
	for (int i = 0; i < lenth; i++)
	{
		cin >> num;
		cur->next=new ListNode(num);
		cur=cur->next;
	}
	l3=LL_merge(l1->next,l2->next);
	while(l3){
		
		cout<data<<" ";
		l3=l3->next;
	}
	return 0;
}