C++ map与set封装实现过程讲解
作者:平凡的人1
一、前情回顾
set 参数只有 key,但是map除了key还有value。我们还是需要KV模型的红黑树的:
#pragma once #include <iostream> #include <assert.h> #include <time.h> using namespace std; enum Color { RED, BLACK, }; template<class K, class V > struct RBTreeNode { pair<K, V> _kv; RBTreeNode<K, V>* _left; RBTreeNode<K, V>* _right; RBTreeNode<K, V>* _parent; Color _col; RBTreeNode(const pair<K,V>& kv) :_kv(kv) ,_left(nullptr) ,_right(nullptr) ,_parent(nullptr) ,_col(RED) {} }; template<class K,class V> class RBTree { typedef RBTreeNode<K, V> Node; public: bool Insert(const pair<K, V>& kv) { if (_root == nullptr) { _root = new Node(kv); _root->_col = BLACK; return true; } Node* parent = nullptr; Node* cur = _root; while (cur) { if (cur->_kv.first < kv.first) { parent = cur; cur = cur->_right; } else if (cur->_kv.first > kv.first) { parent = cur; cur = cur->_left; } else { return false; } } cur = new Node(kv); cur->_col = RED; if (parent->_kv.first < kv.first) { parent->_right = cur; cur->_parent = parent; } else { parent->_left = cur; cur->_parent = parent; } while (parent && parent->_col == RED) { Node* grandfater = parent->_parent; if (parent == grandfater->_left) { Node* uncle = grandfater->_right; //情况一:u存在且为红 if (uncle && uncle->_col == RED) { parent->_col = uncle->_col = BLACK; grandfater->_col = RED; //向上调整 cur = grandfater; parent = cur->_parent; } else { //情况2 if (cur == parent->_left) { RotateR(grandfater); parent->_col = BLACK; grandfater->_col = RED; } //情况3 else { // g // p // c RotateL(parent); RotateR(grandfater); cur->_col = BLACK; grandfater->_col = RED; } break; } } else//parent==grandfater->_right { Node* uncle = grandfater->_left; //情况1:u存在且为红色 if (uncle && uncle->_col == RED) { uncle->_col = parent->_col = BLACK; grandfater->_col = RED; //向上调整 cur = grandfater; parent = cur->_parent; } else { //情况2:u不存在/u存在为黑色 //g // p // c if (cur == parent->_right) { RotateL(grandfater); grandfater->_col = RED; parent->_col = BLACK; } //情况3 // g // p // c else { RotateR(parent); RotateL(grandfater); cur->_col = BLACK; grandfater->_col = RED; } break; } } } //根变黑 _root->_col = BLACK; return true; } void RotateL(Node* parent) { Node* subR = parent->_right; Node* subRL = subR->_left; parent->_right = subRL; if (subRL) subRL->_parent = parent; Node* ppNode = parent->_parent; subR->_left = parent; parent->_parent = subR; if (ppNode == nullptr) { _root = subR; _root->_parent = nullptr; } else { if (ppNode->_left == parent) { ppNode->_left = subR; } else { ppNode->_right = subR; } subR->_parent = ppNode; } } void RotateR(Node* parent) { Node* subL = parent->_left; Node* subLR = subL->_right; parent->_left = subLR; if (subLR) subLR->_parent = parent; Node* ppNode = parent->_parent; parent->_parent = subL; subL->_right = parent; if (ppNode == nullptr) { _root = subL; _root->_parent = nullptr; } else { if (ppNode->_left == parent) { ppNode->_left = subL; } else { ppNode->_right = subL; } subL->_parent = ppNode; } } void InOrder() { _InOrder(_root); } void _InOrder(Node* root) { if (root == nullptr) return; _InOrder(root->_left); cout << root->_kv.first << ":" << root->_kv.second << endl; _InOrder(root->_right); } bool Check(Node*root,int blackNum,int ref) { if (root == nullptr) { //cout << blackNum << endl; if (blackNum != ref) { cout << "违反规则:本条路径的黑色结点的数量根最左路径不相等" << endl; return false; } return true; } if (root->_col == RED && root->_parent->_col == RED) { cout << "违反规则:出现连续的红色结点" << endl; return false; } if (root->_col == BLACK) { ++blackNum; } return Check(root->_left,blackNum,ref) && Check(root->_right,blackNum,ref); } bool IsBalance() { if (_root == nullptr) { return true; } if (_root->_col != BLACK) { return false; } int ref = 0; Node* left = _root; while (left) { if (left->_col == BLACK) { ++ref; } left = left->_left; } return Check(_root,0,ref); } private: Node* _root = nullptr; };
二、简化源码
翻开源码一看
RBTree的结构源码:是KV结构的红黑树
RBTree是通过传入的Value的值来判断类型,也就是一棵泛型的RBTree,通过不同的实例化,实现出了Map和Set:
对于map:传key,对于set:传pair
map的结构简化源码:
set的结构简化源码:
为了让我们的红黑树能够识别set与map我们增加一个模板参数T:
template<class K, class T> class RBTree
对于T模板参数可能是键值Key,也可能是由Key和Value共同构成的键值对。
如果是set容器,那么它传入底层红黑树的模板参数就是Key和Key:
template<class K> class set { private: RBTree<K,K> _t; };
如果是map容器,传入底层红黑树的模板参数就是Key和Key和value的键值对:
class map { private: RBTree<K, pair<const K,V>> _t; };
通过上面,我们可以知道,对于set和map的区别:我们只要通过第二个模板参数就能进行区分,那是不是第一个模板参数就没有意义了呢?
对于insert(const Value&v)来说,需要放入存入的值,确实是这个样子的,插入的值是value,对于set就是key,对于map就是pair。
但是对于find(const Key&key)来说,查找的参数不是value,找的不是pair而是Key,对于map容器来说就不行了。
**红黑树的节点**:set容器:K和T都是键值Key; map容器:K是键值Key,T由Key和Value构成的键值对;但是底层红黑树并不知道上层容器到底是map还是set,因此红黑树的结点当中直接存储T就行了,如果是set的时候,结点当中存储的是键值Key;如果是map的时候,结点当中存储的就是键值对,所以红黑树的结点定义如下,由T类型来决定红黑树存的是key还是pair:
template<class T> //三叉链结构 struct RBTreeNode { T _data; RBTreeNode<T>* _left; RBTreeNode<T>* _right; RBTreeNode<T>* _parent; Color _col; RBTreeNode(const T& data) :_data(data) , _left(nullptr) , _right(nullptr) , _parent(nullptr) , _col(RED) {} };
三、仿函数
这里存在一个问题📝:插入的时候data的大小如何去进行比较:我们并不知道是什么类型是key,还是pair的比较,而我们刚开始kv结构就直接用kv.first去比较了。
对于set是Key
,可以比较
对于map是pair
,那我们要取其中的first来比较,但是pair的大小并不是直接按照first去进行比较的,而我们只需要按照first去进行比较
由于底层的红黑树不知道传的是map还是set容器,当需要进行两个结点键值的比较时,底层红黑树传入的仿函数来获取键值Key,进行两个结点键值的比较:这个时候我们就需要仿函数了,如果是set那就是用于返回T当中的键值Key,如果是map那就是用于返回pair的first:
仿函数/函数对象也是类,是一个类对象。仿函数要重载operator()。
namespace HWC { template<class K,class V> class map { struct MapKeyOfT { const K& operator()(const pair<const K, V>& kv) { return kv.first; } }; public: private: RBTree<K, pair<const K,V>,MapKeyOfT> _t; };
namespace HWC { template<class K> class set { struct SetKeyOfT { const K& operator()(const K& key) { return key; } }; private: RBTree<K,K,SetKeyOfT> _t; };
博主画了个图更加容易进行比对
查找过程,此时就可以套上我们所写的仿函数对象去进行数据的大小比较了:
KeyOfT kot;//仿函数对象 Node* parent = nullptr; Node* cur = _root; while (cur) { if (kot(cur->_data)<kot(data)) { parent = cur; cur = cur->_right; } else if (kot(cur->_data)>kot(data)) { parent = cur; cur = cur->_left; } else { return false; } }
四、迭代器
红黑树的正向迭代器是对结点指针进行了封装,所以这里的正向迭代器就只有一个成员变量:结点的指针,并没有什么其他的地方,迭代器的定义:
template<class T,class Ref,class Ptr> struct __RBTreeIterator { typedef RBTreeNode<T> Node; typedef __RBTreeIterator<T,Ref,Ptr> Self; typedef __RBTreeIterator<T, T&, T*> iterator; Node* _node; __RBTreeIterator(Node*node) :_node(node) {} //普通迭代器的时候,它是拷贝构造 //const迭代器的时候,它是构造,支持用普通迭代器构造const迭代器 __RBTreeIterator(const iterator& s) :_node(s._node) {} }
*:解引用操作,返回对应结点数据的引用:
Ref operator*() { return _node->_data; }
->:成员访问操作符,返回结点数据的引用:
Ptr operator->() { return &_node->_data; }
!=、==:比较简单
bool operator !=(const Self & s) const { return _node != s._node; } bool operator ==(const Self& s) const { return _node == s._node; }
这里的迭代器重点是迭代器的++:
一个结点的正向迭代器进行++
操作后,根据红黑树中序(左、根、右)找到当前结点的下一个结点,中序的第一个节点是最左,迭代器的++
怎么去找:
如果节点的右子树不为空,++
就是找右子树的最左节点
如果节点的右子树为空,++
就是找祖先(孩子是父亲的左的那个祖先)
代码实现:
Self& operator++() { if (_node->_right) { Node* min = _node->_right; while (min->_left) { min = min->_left; } _node = min; } else { Node* cur = _node; Node* parent = cur->_parent; while (parent && cur == parent->_right) { cur = cur->_parent; parent = parent->_parent; } _node = parent; } return *this; }
迭代器的--
对于–,如果是根,–就是左子树,找到左子树最大的那一个(最右节点)
如果节点的左子树不为空,--
找左子树最右的节点
如果节点的左子树为空,--
找祖先(孩子是父亲的右的祖先)
代码实现:
Self& operator--() { if (_node->_left) { Node* max = _node->_left; while (max->_right) { max = max->_right; } _node = max; } else { Node* cur = _node; Node* parent = cur->_parent; while (parent&&cur==parent->_left) { cur = cur->_parent; parent = parent->_parent; } _node = parent; } return *this; }
不要忘记迭代器的两个核心成员:begin()与end()
begin()
:返回中序(左、根、右)第一个结点的正向迭代器,即最左节点,返回的是最左节点,直接找最左节点即可
end()
:返回中序(左、根、右)最后一个结点下一个位置的正向迭代器,这里直接用空指针
template<class K, class T,class KeyOfT> class RBTree { typedef RBTreeNode<T> Node; public: typedef __RBTreeIterator<T> iterator; iterator begin() { Node* left = _root; while (left && left->_left) { left = left->_left; } return iterator(left); } iterator end() { return iterator(nullptr); } }
五、set的实现
通过前面底层红黑树的接口进行套用即可实现set的实现:
值得注意的是🔴:typename:没有实例化的模板,区分不了是静态变量还是类型,typename告诉编译器是类型
#pragma once #include "RBTree.h" namespace hwc { template <class K> class set { struct SetKeyOfT { const K& operator()(const K& key) { return key; } }; public: //typename:没有实例化的模板,区分不了是静态变量还是类型,typename告诉编译器是类型 typedef typename RBTree<K, K, SetKeyOfT>::const_iterator iterator;//key不可以修改 typedef typename RBTree<K, K, SetKeyOfT>::const_iterator const_iterator; iterator begin() const { return _t.begin(); } iterator end() const { return _t.end(); } pair<iterator,bool> insert(const K& key) { //底层红黑树的iterator是普通迭代器 pair<typename RBTree<K, K, SetKeyOfT>::iterator, bool> ret = _t.Insert(key); return pair<iterator, bool>(ret.first, ret.second);//用普通迭代器构造const迭代器 } private: RBTree<K, K,SetKeyOfT> _t; }; void test_set() { int a[] = { 4, 2, 6, 1, 3, 5, 15, 7, 16, 14 }; set<int> s; for (auto e : a) { s.insert(e); } set<int>::iterator it = s.begin(); while (it != s.end()) { cout << *it << " "; ++it; } cout << endl; for (auto e : s) { cout << e << " "; } cout << endl; } }
六、map的实现
同样是套用上底层红黑树的接口,不过map的实现有一个很重要的地方,那就是[]的实现
#pragma once #include "RBTree.h" namespace hwc { template<class K,class V> class map { struct MapkeyOfT { const K& operator()(const pair<const K, V>& kv) { return kv.first; } }; public: //typename:没有实例化的模板,区分不了是静态变量还是类型,typename告诉编译器是类型 typedef typename RBTree<K, pair<const K, V>, MapkeyOfT>::iterator iterator; typedef typename RBTree<K, pair<const K, V>, MapkeyOfT>::const_iterator const_iterator; iterator begin() { return _t.begin(); } iterator end() { return _t.end(); } const_iterator begin() const { return _t.begin(); } const_iterator end() const { return _t.end(); } pair<iterator,bool> insert(const pair<const K, V>& kv) { return _t.Insert(kv); } V& operator[](const K& key) { pair<iterator, bool> ret = insert(make_pair(key, V())); return ret.first->second; } private: RBTree<K, pair<const K, V>, MapkeyOfT> _t; }; void test_map() { int a[] = { 4, 2, 6, 1, 3, 5, 15, 7, 16, 14 }; map<int, int> m; for (auto e : a) { m.insert(make_pair(e, e)); } map<int, int>::iterator it = m.begin(); while(it!=m.end()) { it->second++; cout << it->first << ":" << it->second << endl; ++it; } cout << endl; map<string, int> countMap; string arr[] = { "苹果","西瓜","香蕉","苹果"}; for (auto& e : arr) { countMap[e]++; } for (auto& kv : countMap) { cout << kv.first << ":" << kv.second << endl; } } }
七、红黑树代码
最后,在这里送上源码:
#pragma once #pragma once #include <iostream> #include <assert.h> #include <time.h> using namespace std; enum Color { RED, BLACK, }; template<class T> struct RBTreeNode { T _data; RBTreeNode<T>* _left; RBTreeNode<T>* _right; RBTreeNode<T>* _parent; Color _col; RBTreeNode(const T& data) :_data(data) , _left(nullptr) , _right(nullptr) , _parent(nullptr) , _col(RED) {} }; template<class T,class Ref,class Ptr> struct __RBTreeIterator { typedef RBTreeNode<T> Node; typedef __RBTreeIterator<T,Ref,Ptr> Self; typedef __RBTreeIterator<T, T&, T*> iterator; Node* _node; __RBTreeIterator(Node*node) :_node(node) {} //普通迭代器的时候,它是拷贝构造 //const迭代器的时候,它是构造,支持用普通迭代器构造const迭代器 __RBTreeIterator(const iterator& s) :_node(s._node) {} Ref operator*() { return _node->_data; } Ptr operator->() { return &_node->_data; } Self& operator++() { if (_node->_right) { Node* min = _node->_right; while (min->_left) { min = min->_left; } _node = min; } else { Node* cur = _node; Node* parent = cur->_parent; while (parent && cur == parent->_right) { cur = cur->_parent; parent = parent->_parent; } _node = parent; } return *this; } Self& operator--() { if (_node->_left) { Node* max = _node->_left; while (max->_right) { max = max->_right; } _node = max; } else { Node* cur = _node; Node* parent = cur->_parent; while (parent&&cur==parent->_left) { cur = cur->_parent; parent = parent->_parent; } _node = parent; } return *this; } bool operator !=(const Self & s) const { return _node != s._node; } bool operator ==(const Self& s) const { return _node == s._node; } }; template<class K, class T,class KeyOfT> class RBTree { typedef RBTreeNode<T> Node; public: typedef __RBTreeIterator<T,T&,T*> iterator; typedef __RBTreeIterator<T,const T&,const T*> const_iterator; const_iterator begin() const { Node* left = _root; while (left && left->_left) { left = left->_left; } return const_iterator(left); } const_iterator end() const { return const_iterator(nullptr); } iterator begin() { Node* left = _root; while (left && left->_left) { left = left->_left; } return iterator(left); } iterator end() { return iterator(nullptr); } pair<iterator,bool> Insert(const T& data) { if (_root == nullptr) { _root = new Node(data); _root->_col = BLACK; return make_pair(iterator(_root),true); } KeyOfT kot; Node* parent = nullptr; Node* cur = _root; while (cur) { if (kot(cur->_data) < kot(data)) { parent = cur; cur = cur->_right; } else if (kot(cur->_data) > kot(data)) { parent = cur; cur = cur->_left; } else { return make_pair(iterator(cur),false); } } cur = new Node(data); Node* newnode = cur; cur->_col = RED; if (kot(parent->_data) < kot(data)) { parent->_right = cur; cur->_parent = parent; } else { parent->_left = cur; cur->_parent = parent; } while (parent && parent->_col == RED) { Node* grandfater = parent->_parent; if (parent == grandfater->_left) { Node* uncle = grandfater->_right; //情况一:u存在且为红 if (uncle && uncle->_col == RED) { parent->_col = uncle->_col = BLACK; grandfater->_col = RED; //向上调整 cur = grandfater; parent = cur->_parent; } else { //情况2 if (cur == parent->_left) { RotateR(grandfater); parent->_col = BLACK; grandfater->_col = RED; } //情况3 else { // g // p // c RotateL(parent); RotateR(grandfater); cur->_col = BLACK; grandfater->_col = RED; } break; } } else//parent==grandfater->_right { Node* uncle = grandfater->_left; //情况1:u存在且为红色 if (uncle && uncle->_col == RED) { uncle->_col = parent->_col = BLACK; grandfater->_col = RED; //向上调整 cur = grandfater; parent = cur->_parent; } else { //情况2:u不存在/u存在为黑色 //g // p // c if (cur == parent->_right) { RotateL(grandfater); grandfater->_col = RED; parent->_col = BLACK; } //情况3 // g // p // c else { RotateR(parent); RotateL(grandfater); cur->_col = BLACK; grandfater->_col = RED; } break; } } } //根变黑 _root->_col = BLACK; return make_pair(iterator(newnode),true); } void RotateL(Node* parent) { Node* subR = parent->_right; Node* subRL = subR->_left; parent->_right = subRL; if (subRL) subRL->_parent = parent; Node* ppNode = parent->_parent; subR->_left = parent; parent->_parent = subR; if (ppNode == nullptr) { _root = subR; _root->_parent = nullptr; } else { if (ppNode->_left == parent) { ppNode->_left = subR; } else { ppNode->_right = subR; } subR->_parent = ppNode; } } void RotateR(Node* parent) { Node* subL = parent->_left; Node* subLR = subL->_right; parent->_left = subLR; if (subLR) subLR->_parent = parent; Node* ppNode = parent->_parent; parent->_parent = subL; subL->_right = parent; if (ppNode == nullptr) { _root = subL; _root->_parent = nullptr; } else { if (ppNode->_left == parent) { ppNode->_left = subL; } else { ppNode->_right = subL; } subL->_parent = ppNode; } } void InOrder() { _InOrder(_root); } void _InOrder(Node* root) { if (root == nullptr) return; _InOrder(root->_left); cout << root->_kv.first << ":" << root->_kv.second << endl; _InOrder(root->_right); } bool Check(Node* root, int blackNum, int ref) { if (root == nullptr) { //cout << blackNum << endl; if (blackNum != ref) { cout << "违反规则:本条路径的黑色结点的数量根最左路径不相等" << endl; return false; } return true; } if (root->_col == RED && root->_parent->_col == RED) { cout << "违反规则:出现连续的红色结点" << endl; return false; } if (root->_col == BLACK) { ++blackNum; } return Check(root->_left, blackNum, ref) && Check(root->_right, blackNum, ref); } bool IsBalance() { if (_root == nullptr) { return true; } if (_root->_col != BLACK) { return false; } int ref = 0; Node* left = _root; while (left) { if (left->_col == BLACK) { ++ref; } left = left->_left; } return Check(_root, 0, ref); } private: Node* _root = nullptr; };
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