面试金典--删除未排序链表重复节点

编写代码，删除未排序链表的重复节点

思路：

hash记录当前节点是否出现过

#include <iostream>
#include <string>
#include <fstream>
#include <map>
#include <algorithm>
#include <vector>
#include <ctime>
#include <bitset>

using namespace std;

template<typename T>
class Node
{
public:
    Node<T> *next;
    T data;

    Node(T d):data(d),next(NULL){}
    void appendToTail(T d)
    {
        Node<T> *end = new Node<T>(d);
        Node<T> *n = this;
        while(n->next != NULL)
        {
            n = n->next;
        }
        n->next = end;
    }
};

int main()
{
    Node<int> *head = new Node<int>(1);
    int i;
    for(i = 2 ; i < 6 ; ++i)
    {
        head->appendToTail(i);
    }
    //2.1
    map<int,bool> nodesInList;
    head->appendToTail(2);
    map<int,bool>::iterator it;
    Node<int> *prev = NULL;
    Node<int> *headOrg = head;
    while(head != NULL)
    {
        it = nodesInList.find(head->data);
        if(it != nodesInList.end())
        {
            prev->next = head->next;
        }
        else
        {
            nodesInList[head->data] = true;
            prev = head;
        }
        head = head->next;
    }
    while(headOrg != NULL)
    {
        cout<<headOrg->data<<endl;
        headOrg = headOrg->next;
    }
    return 0;
}

扩展：没有缓冲区？那么只能O(N^2)了。上面相当于空间换时间