Question

最终，程序将按字母顺序打印出名称列表，以及与该名称相关的其他属性。换句话说，输出屏幕将如下所示：

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

同样，名字在此列表中按字母顺序排列，但属性与它们一起打印。关于我的int main（），我不知道如何开始对这些名称进行排序并将它们整理好。我有一个必须排序的未排序列表（使用按正确顺序添加/插入这些名称的函数）。

    //
    // 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

我假设这是作业，如果不是，简单的答案就是你应该在那个容器中使用std::list和sort方法。

main中要放什么？最有可能的是：sort_list( newlist )，其中指针通过引用传递（因为列表的头部可能会改变）。至于如何实现它，它取决于你想要实现的排序算法，最简单的可能是冒泡排序，列表的下一个最佳选择是合并排序。谷歌为他们，如果您需要有关算法的帮助，请回来询问。

与此同时，您可能希望处理我提出的问题的问题：代码格式化，内存泄漏（如果找不到位置，则在插入时，以及程序结束时），正确性任何时候的数据结构...我还没有进行深入分析，但我感觉你的add_after_find可能会失败，因为它需要在列表的尾部添加一个元素...在你之前开始考虑排序，你应该确保输入是正确的。与开始添加更多代码相比，调试当前问题更容易。

我正在创建一个双向链表，它将按字母顺序排列名称列表，但我不确定要在int main（）函数中放入什么

1 个答案: