一、list的使用

list实质上就是双向链表，用法和vector、string差不多

int main()
{
	list<int> lt1;
	lt1.push_back(1);
	lt1.push_back(2);
	lt1.push_back(3);
	lt1.push_front(0);
 
	//遍历
 
	//迭代器
	list<int>::iterator it = lt1.begin();
	//这里用"!=",是因为end位置是在头哨兵位
	while (it != lt1.end())
	{
		cout << *it << " ";
		it++;
	}
	cout << endl;
 
 
	//范围for
	for (auto e : lt1)
	{
		cout << e << " ";
	}
	cout << endl;
 
	//反向迭代器
	//list::reverse_iterator rit = lt1.rbegin();
	auto rit = lt1.rbegin();
	while (rit != lt1.rend())
	{
		cout << *rit << " ";
		rit++;
	}
	cout << endl;
 
	lt1.pop_front();
	lt1.pop_back();
 
	for (auto e : lt1)
	{
		cout << e << " ";
	}
	cout << endl;
}

 二、list的模拟实现

namespace Cris
{
	//节点
	template<class T>
	struct list_node
	{
		list_node* _next;
		list_node* _prev;
		T _data;
 
		list_node(const T& val = T())
			:_next(nullptr)
			,_prev(nullptr)
			,_data(val)
		{}
	};
 
 
	//迭代器
 
	// typedef __list_iterator iterator;
	// typedef __list_iterator const_iterator;
	template<class T, class Ref, class Ptr>
	struct _list_iterator
	{
		typedef list_node Node;
		typedef _list_iterator self;
		Node* _node;
 
		_list_node(Node* node)
			:_node(node)
		{}
 
		//析构函数 -- 节点不属于迭代器，不需要迭代器释放
		//拷贝构造和赋值重载 -- 默认生成的浅拷贝
 
		Ref operator* ()
		{
			return _node->_data;
		}
 
		Ptr operator->()
		{
			return &_node->_data;
		}
 
		self& operator++()
		{
			_node = _node->_next;
			return *this;
		}
 
		self operator++(int)
		{
			self tmp(*this);
			_node = _node->_next;
			return tmp;
		}
 
		self& operator--()
		{
			_node = _node->_prev;
			return *this;
		}
 
		self operator--(int)
		{
			self tmp(*this);
			_node = _node->_prev;
			return tmp;
		}
 
		bool operator != (const self& it)
		{
			return _node != it._node;
		}
 
		bool operator == (const self& it)
		{
			return _node == it._node;
		}
	};
 
	//链表
	template<class T>
	class list
	{
		typedef list_node<int> Node;
	public:
		typedef _list_node iterator;
		typedef _list_nodeconst T&, const T*> const_iterator;
 
		const_iterator begin() const
		{
			return const_iterator(_head->next);
		}
 
		const_iterator end() const
		{
			return const_iterator(_head);
		}
 
		iterator begin()
		{
			return iterator(_head->next);
		}
 
		iterator end()
		{
			return iterator(_head);
		}
 
		list()
		{
			_head = new Node();
			_hend->next = _head;
			_head->_prev = _head;
		}
 
		void empty_init()
		{
			_head = new Node();
			_head->_next = _head;
			_head->_prev = _head;
		}
 
		template <class InputIterator>
		list(InputIterator first, InputIterator last)
		{
			empty_init();
 
			while (first != last)
			{
				push_back(*first);
				++first;
			}
		}
 
		void swap(list& lt)
		{
			std::swap(_head, lt._head);
		}
 
		//lt2(lt1)    现代写法
		list(const list& lt)
		{
			empty_init();
			list tmp(lt.begin(), lt.end());
			swap(tmp);
		}
 
		//lt2 = lt1
		list& operator=(list lt)
		{
			swap(lt);
			return *this;
		}
 
		~list()
		{
			clear();
			delete _head;
			_head = nullptr;
		}
 
		void clear()
		{
			iterator it = begin();
			while (it != end())
			{
				it = erase(it);
			}
		}
 
		void push_back(const T& x)
		{
			/*Node* tail = _head->_prev;
			Node* newnode = new Node(x);
			tail->_next = newnode;
			newnode->_prev = tail;
			newnode->_next = _head;
			_head->_prev = newnode;*/
 
			insert(end(), x);
		}
 
		void push_front(const T& x)
		{
			insert(begin(), x);
		}
 
		void pop_back()
		{
			//end() 是头哨兵位，--就是list的尾
			erase(--end());
		}
 
		void pop_front()
		{
			erase(begin());
		}
 
		//插入在pos位置之前
		iterator insert(iterator pos, const T& x)
		{
			Node* newNode = new Node(x);
			Node* cur = pos._node;
			Node* prev = cur->_prev;
 
			prev->_next = newNode;
			newNode->_prev = prev;
			newNode->_next = cur;
			cur->_prev = newNode;
 
			return iterator(newNode);
		}
 
		iterator erase(iterator pos)
		{
			assert(pos != end());
 
			Node* cur = pos._node;
			Node* prev = cur->_prev;
			Node* next = cur->_next;
 
			prev->_next = next;
			next->_prev = prev;
			delete cur;
 
			return iterator(next);
		}
 
	private:
		Node* _head;
 
	};
}

迭代器失效问题：

因为list的底层结构为带头结点的双向循环链表，因此在list中进行插入时是不会导致list的迭代器失效的，只有在删除时才会失效，并且失效的只是指向被删除节点的迭代器，其他迭代器不会受到影响。

int main()
{
	int arr[] = { 1,2,3,4,5,6,7,8,9 };
	Cris::list<int> lt(arr, arr + (sizeof(arr) / sizeof(arr[0])));
 
	auto it = lt.begin();
	while (it != lt.end())
	{
		//erase()函数执行后，it所指向的节点已被删除
		//因此it无效，在下一次使用it时，必须先给其赋值
		it = lt.erase(it);
	}
 
	return 0;
}

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