我正在創建一個雙向鏈接列表，該列表將按字母順序排列一個名稱列表，但是我不確定要在int main（）函數中放入什么

Question

最終，該程序將按字母順序打印出名稱列表，以及與該名稱關聯的其他屬性。 換句話說，輸出屏幕將顯示如下：

Ares: Greek, fire, sword.
Freia: Norse, water, bow and arrow.
Poseidon: Greek, horses, ocean.
Thor: Norse, chariot, hammer.
Zeus: Greek, cloud, lightning.

同樣，名字在此列表中按字母順序排列，但屬性與它們並列顯示。 關於我的int main（），我不確定應該如何開始對這些名稱進行排序和排序。 我有一個必須排序的未排序列表（使用將這些名稱添加/插入正確順序的函數）。

    //
    // This is a standard library support code to the chapters of the book
    // "Programming -- Principles and Practice Using C++" by Bjarne Stroustrup
    //

    #ifndef STD_LIB_FACILITIES_GUARD
    #define STD_LIB_FACILITIES_GUARD 1

    #include <cmath>
    #include <iostream>
    #include <vector>
    #include <stdexcept>
    #include <string>

    using namespace std;

    //------------------------------------------------------------------------------

    // Helper function to show an error message
    inline void error(const string& errormessage)
    {
    throw runtime_error(errormessage);
    }

    //------------------------------------------------------------------------------

    #endif // STD_LIB_FACILITIES_GUARD

    //------------------------------------------------------------------------------

    struct Link {
    string name;
    string mythology;
     string vehicle;
    string weapon;

    Link* prev;
    Link* succ;
    Link(const string& n, const string& a, const string& b, const string&c,Link* p = 0,    
    Link* s = 0)
: name(n), mythology(a), vehicle(b), weapon(c), prev(p), succ(s) { }
    };



    Link* insert(Link* p, Link* n)    // insert n before p; return n
    {
    if (n==0) return p;
    if (p==0) return n;
     n->succ = p;        // p comes after n
    if (p->prev) p->prev->succ = n;
     n->prev = p->prev;    // p's predecessor becomes n's predecessor
     p->prev = n;        // n becomes p's predecessor
     return n;
     }

    void print_all(Link *p)
    {
Link *current;
current = p;
while(current)
{
    cout<<"For this link we have: \n";
    cout<<"Name: "<<current->name<<".\n";
    cout<<"Info1: "<<current->mythology<<".\n";
    cout<<"Info2: "<<current->vehicle<<".\n";
    cout<<"Info3: "<<current->weapon<<".\n";
    current = current->succ;
}

    }
    Link * add_after_find(Link *p, Link *n,const string& s )
     {   Link *current = 0;
current = p;
/* empty list */
if(p == 0)
{   cout<<"List is empty so string not found so not added after it. \n";

    return  0;
}
/*  DO WE NEED ONE LINK ONLY */
else if(p->succ == 0)   /* one link only */
{
    if(p->name == s)
    {

        /* add after link with s */
        /* p in front */
        p->succ = n;
        n->prev = p;
        p->prev = 0;
        n->succ = 0;

        return p;      
    }   /* end of if names =  */
    else {
        cout<<"String not found in link listed so not added. \n";
        return p;

    }

}  /* end of one link */

else /* two or more links */

{   
    current = p;
    while(current->succ)
    {
        if (s == current->name)


        {

            /* then n goes AFTER this link */
            n->prev = current;
            n->succ = current->succ;
            current->succ = n;


            return p;
        }  /* end of name matches */

        else 
        {
            current = current->succ;
        }
    }// end of while
    /* if outside of while then we are at last link with a current -> name 
     so s not found  */
    cout<<"String is not found so not add after it. \n";
    return p;
}  // end of else 2 or more
}  // end of function

---

int main()
{      
    Link*newlist = new Link("Thor","Norse","chariot","hammer");
    newlist = add_after_find(newlist,new Link("Hera","Greek", "horse", "arrow"),"Thor");
    newlist = add_after_find(newlist,new Link("Poseidon","Greek", "ocean", "trident"),"Freia");
     newlist = add_after_find(newlist,new Link("Ares","Greek", "fire", "sword"),"Poseidon");
     newlist = add_after_find(newlist,new Link("Zeus","Greek", "cloud", "lightning"),"Ares");

    print_all(newlist);
    cout<<"Now let's alphabetize these five gods.\n";
    system("Pause");
    return 0;
}

Answer 1

我假設這是家庭作業，如果不是，簡單的答案是您應該在該容器中使用std::list和sort方法。

main什么？ 最有可能在以下代碼行上進行sort_list( newlist ) ： sort_list( newlist ) ，其中指針通過引用傳遞（因為列表的頭部可能會更改）。 至於如何實現，取決於您要實現的排序算法，最簡單的可能是冒泡排序，列表的次佳選擇是合並排序。 Google，如果您需要有關算法的幫助，請回來詢問。

同時，您可能需要處理我作為對該問題的評論而提出的問題：代碼格式化，內存泄漏（在插入時（如果找不到位置，則在插入時以及在程序末尾）），數據結構的正確性在任何時候……我都沒有做深入的分析，但是我有一種感覺，當您的add_after_find需要在列表的add_after_find添加元素時，它可能會失敗。甚至在考慮進行排序之前，您都應該確保輸入正確。 比起開始添加更多代碼，調試當前問題要容易得多。