[数据结构与算法]单链表

#include<iostream>
using namespace std;
struct Node
{
	int date;//数据域
	struct Node* pNext;//指针域
};
class List
{
public:

	int len;//用来存放有效结点的个数
	int i;
	int val;//用来临时存放用户输入的结点的值
	struct Node* creat_list();//创建
	bool is_empty(struct Node* pHead);//是否为空
	int length_list(struct Node*);//长度
	bool insert_list(struct Node*,int,int);//插入
	bool delete_list(struct Node*, int, int *);//删除
	void scort_list(struct Node*);//排序
private:
	Node* pHead;
};

struct Node* create_list()
{
	
	if (NULL == pHead)
	{
		cout << "分配失败" << endl;
	}

	cout << "请输入需要生成链表结点的个数：len= ";

	for (int i = 0; i < List::len; ++i)
	{
		cout << "请输入第几个结点的值：";
		i += 1;
	}

};

bool is_empty(struct Node* pHead)//是否为空
{
	if (NULL == pHead->pNext)
		return true;
	else
		return false;
};
int length_list(struct Node*pHead)//长度
{
	struct Node* p = pHead->pNext;
	int len = 0;
	while (p)
	{
		++len;
		p = p->pNext;
	}
	return len;
};
void soert_list(struct Node*)//排序
{
	int i, j, t;
	int len = length_list(pHead);
	Node* pNext = p, q;
	for ( i = 0,p=pHead->pNext; i < List::len - 1; ++i,p=p->pNext)
	{
		for (j = i + 1,q=p->pNext; j < len; ++j,q=q->pNext)
		{
			if (p->date > q->date)//类似于数组中的：a[i]>a[j]
			{
				t = p->date;// t = a[i];
				p->date = q->date;//a[i] = a[j];
				q->date = t;//a[j] = t;
			}
		}
	}
	return;
};

bool insert_list(struct Node*, int pos, int val)//在链表第pos个结点前插入一个新结点，值为val
{
	int i = 0;
	struct Node* p = List::pHead;
	while (NULL != p && i < pos - 1)
	{
		p = p->pNext;
		++i;
	}
	if (i > pos - 1 || NULL == p->pNext)
		return false;
	struct Node* q = p->pNext;
	val = q->date;

};
bool delete_list(struct Node*, int, int*)//删除,指向链表，删除结点，删除的结点放到第三个结点里去
{
	int i = 0;
	struct Node* p = List::pHead;
	while (NULL != p && i < pos - 1)
	{
		p = p->pNext;
		++i;
	}
	if (i > pos - 1 || NULL == p->pNext)
		return false;
	struct Node* q = p->pNext;
	*pVal = q->date;
	p->pNext = p->pNext->pNext;
	free(q);
	q = NULL;
};
int main()
{
	struct Node* pHead = NULL;
	pHead = creak_list();
	int len = length_list(pHead);
	cout << "链表的长度为" << len << endl;

	return 0;
}