单链表的c++实现，使用模板

来源：互联网发布：数据运营编辑：程序博客网时间：2024/05/18 00:27

节点的构造

#include<iostream>using namespace std;template<class _Ty>class list;template<class _Ty>class list_node{friend class list<_Ty>;public:list_node():next(NULL){}list_node(const _Ty item, list_node<_Ty>* tail=NULL):data(item),next(tail){}~list_node(){}private:_Ty data;list_node *next;};

//list方法实现

#include "list_node.h"template<class _Ty>class list{public:list():head(new list_node<_Ty>()){}~list(){delete head;}public:void Insert(const _Ty x,int i)//将数据x插入到位置i之后{list_node<_Ty> *phead = head;list_node<_Ty> *p = new list_node<_Ty>(x);while(--i && phead){phead = phead->next;}p->data = x;p->next = phead->next;phead->next = p;}void Show()const{list_node<_Ty> *phead = head->next;while(phead != NULL){cout<<phead->data<<"->";phead = phead->next;}cout<<endl;}_Ty Get(int i)  //查找某个位置的节点的值{list_node<_Ty> *p = head->next;while(--i){p = p->next;}return p->data;}_Ty Remove(int i)  //删除i位置的节点，返回被删除节点的data。{list_node<_Ty>* p = head;list_node<_Ty> *q;while(--i && p){p = p->next;}q = p->next;p->next = q->next;_Ty tmp = q->data;delete q;return tmp;}void RemoveAll(_Ty item)  //删除所有p->data==item的节点{list_node<_Ty> *p = head->next;int i = 1;while(p != NULL){if(p->data == item){p = p->next;  //记录p的下一个节点，下次从p->next处开始查询Remove(i);}else{i++;p = p->next;}}}void Clear(){list_node<_Ty>* p = head->next;while(p != NULL){p = p->next;Remove(1);}}size_t size()const  //不加head节点{int i = 0;list_node<_Ty> *p = head->next;while(p != NULL){i++;p = p->next;}return i;}private:list_node<_Ty> *head;};

//测试实例

#include "list.h"void main(){list<int> st;for(int i=1; i<10; ++i)st.Insert(i,i);//st.Insert(5,5);//st.Insert(2,2);//st.Show();//st.Remove(2);//st.Show();//st.RemoveAll(2);cout<<st.Find(4)<<endl;cout<<st.Find(4)<<endl;cout<<st.Find(4)<<endl;//cout<<st.size()<<endl;//cout<<st.Get(4)<<endl;//st.Clear();st.Show();}

面试宝典中---查找单链表的倒数第K个元素

实现思路主要有三个：

1、使用两个指针fast和slow遍历，让fast先遍历K个节点，此时fast和slow同时遍历，当fast遍历到NULL节点时，即为所求 --->（此方法只遍历一次）

_Ty Find2(int K){list_node<_Ty> *fast = head->next;list_node<_Ty> *slow = head->next;while(K--){fast = fast->next;}while(fast != NULL){fast = fast->next;slow = slow->next;}return slow->data;}

2、首先遍历所有节点统计节点个数size，要查找倒数第K个元素，即为查找正数第size-K+1个元素。在重新遍历到suze-K+1位置处即可。

_Ty Find(int K) //找出倒数第K个元素{list_node<_Ty> *p = head->next;int i = 1;while(p->next != NULL){i++;  //统计总的元素个数p = p->next;}  int j = i-K+1;  //相当于查找第j个元素p = head->next;while(--j && p!=NULL){p = p->next;}return p->data;}

3、使用两个指针，一个负责遍历，一个记录遍历前节点的指针；首先遍历K个节点，判断当前节点是否为空，

若为空，则说明遍历前节点（即首节点为所求）；若不为空，记录指针指向下一个节点，赋给遍历指针，遍历指针继续遍历K个节点，依次类推...

_Ty Find3(int K){list_node<_Ty> *p = head->next;list_node<_Ty> *q = head->next;while(p != NULL){while(K--){p = p->next;}if(p == NULL)return q;else{q = q->next;p = q;}}}

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