1

我有输出：

节点 1：鲍勃·乔·吉尔 Jeff Jill

但我希望它在一个名称重复的地方被发送到单链表的前面，所以它会变成

节点 1：吉尔鲍勃乔杰夫

而且我无法实现这一点。

这是我的代码：

string employers[] = {"Jill", "Jeff", "Bob", "Joe", "Monica", "Luis"}; 

struct node {
    node(string name="") {data=name; next=NULL; }

    string data;

    node *next;
    node *prev;
};


class list {
public:
    list(int N=0, int value=0);
    ~list();

    void put(int);
    friend ostream & operator << (ostream &, const list &);

private:
    int N;
    node *head;

};



void list::put(int i) {
    string employee_name = employers[i];
    node * p = new node(g);
    node * pp = head;

    while (pp - > next) {


        pp = pp - > next;
        for (int b=6; b<6; b++) {
           if (p-> data == names[b]
             cout << "found";
    }

    pp - > next = p;

    N++;

}

我遇到的困难是，我将如何比较链接列表中的每个条目？我做了一个节点 *prev，但我不完全确定如何比较节点。

4

2 回答 2

0
  1. 总是写小函数
  2. 如果一个函数看起来更大，总是分解成更小的函数
  3. 尽量避免全局数据，如有必要，尽量传递全局值而不是直接对其进行操作

这是您的解决方案。我添加了一个查找功能并更正了指针管理。

class list {
public:
    list():head(NULL), N(0){}
    ~list(){
    //Implementation for cleanup
     }

void put(int i){ //left this function so that your code wont break but try removing it
  put(employee_names[i]);
}

void put(string name){  //rather than accessing the global data, use the value passed
    node* p = new node(name);
    p->next=p->prev=NULL;
    node* pp = find(name);
    if(pp==NULL){
      // No match found, append to rear
      if(head==NULL)
        head=p;  //list empty, add first element
      else{
        node* cur=head;
        while(cur->next!=NULL) //Keep looking until a slot is found
          cur=cur->next;
        cur->next=p;
        p->prev=cur;
      }
    }
    else{
        //Match found, detach it from its location
        node* pPrev = pp->prev;
        pPrev->next = pp->next;
        pp->next->prev=pPrev;
        p->next = head; //append it to the front & adjust pointers
        head->prev=p;
    }
    N++;
    }

    //MER: finds a matching element and returns the node otherwise returns NULL
    node* find(string name){
        node *cur=head;
        if(cur==NULL) // is it a blank list?
          return NULL;
        else if(cur->data==head) //is first element the same?
          return head;
        else   // Keep looking until the list ends
          while(cur->next!=NULL){
          if(cur->data==name)
            return cur;
            cur=cur->next;
          }
        return NULL;
}
friend ostream& operator << (ostream& os, const list& mylist);

private:
    int N;
    node *head;

};

现在有些人可能会告诉您使用 STL n 中的列表永远不要编写自己的代码，因为您无法击败 STL，但对我来说，您自己实现以清楚了解它在现实中的工作原理是件好事。

于 2013-09-24T05:19:21.130 回答
0

如果这不是学校作业，我会这样做。

class EmployerCollection
{    
public:
    bool AddEmployer(const std::string& name)
    {
        EmployerList::const_iterator it = std::find(m_employers.begin(), m_employers.end(), name);
        if (it != m_employers.end()) // Already exists in list.
        {
            m_employers.splice(m_employers.begin(), m_employers, it, std::next(it));
            return true;
        }
        m_employers.push_front(name);
        return false;
    }

private:
    typedef std::list<std::string> EmployerList;
    EmployerList m_employers;
};

int main()
{
    const int NUM_EMPLOYERS = 15;
    std::string employers[NUM_EMPLOYERS] = {"Jill", "Jeff", "Jill"};
    EmployerCollection c;

    for (int i=0; i<NUM_EMPLOYERS; i++)
    {
        bool duplicate = c.AddEmployer(employers[i]);
        printf("Added %s to employer list - duplicate: %s \n", employers[i].c_str(), duplicate ? "True" : "False");
    }
}
于 2013-09-24T05:54:41.713 回答