c++ - 如何在C++中将对象转换为链表？

Question

在我的 .h 文件中，我有两个函数:

String( const char * s = "");
String( const String & s );

我的目标是将这两个构造函数创建的对象转换为链表。所以我写了一个转换函数:

static ListNode * stringToList(const char *s)
 {
     ListNode * h = new ListNode(s[0], NULL);
     ListNode * c = h;
     for(int i = 0; s[i] != '\0'; ++i)
     {
         c->next = new ListNode(s[i], NULL);
         c = c->next;
     }
     return h;
 }

我知道也许在 char*s 的情况下，我们可以使用

String::String( const char * s)
 {
       ListNode::stringToList(s);
 }

但是在字符串对象的情况下，

String::String( const String & s )

我怎么可能那样做？我对链表有点困惑/此外，此函数之后的所有函数都使用 (const String & s) 参数。在链接列表上操作变得更难，我很困惑如何通过链接列表完成对数据的操作，而调用的所有参数都是“const String & s”。字符串对象

我的整个 link.h 文件:

    #include <iostream>

using namespace std;

class String
{
public:
    /// Both constructors should construct
    /// from the parameter s
     String( const char * s = "");
     String( const String & s );
     String operator = ( const String & s );
     char & operator [] ( const int index );
     int length() const {return ListNode::length(head);}
     int indexOf( char c ) const;
     bool operator == ( const String & s ) const;
     bool operator < ( const String & s ) const;
     /// concatenates this and s
     String operator + ( const String & s ) const;
     /// concatenates s onto end of this
     String operator += ( const String & s );
     void print( ostream & out );
     void read( istream & in );
     ~String();

private:
     bool inBounds( int i )
     {
         return i >= 0 && i < length();
     }

     struct ListNode
     {
     char info;
     ListNode * next;
     ListNode(char newInfo, ListNode * newNext)
         :info( newInfo ), next( newNext )
     {
     }
     // HINT: some primitives you *must* write and use, recursion?
     static ListNode * stringToList(const char *s)
     {
         ListNode * h = new ListNode(s[0], NULL);
         ListNode * c = h;
         for(int i = 0; s[i] != '\0'; ++i){
             c->next = new ListNode(s[i], NULL);
             c = c->next;
         }
         return h;
     }
     static ListNode * copy(ListNode * L)
     {
         return !L ? nullptr : new ListNode(L->info, copy(L->next));
     }
     static bool equal(ListNode * L1, ListNode * L2)
     {
         return (L1->info != L2->info) ? false : equal(L1->next, L2->next);
     }
     static ListNode * concat(ListNode * L1, ListNode * L2)
     {
         return !L1 ? copy(L2) : new ListNode(L1->info, concat(L1->next, L2));
     }
     static int compare(ListNode * L1, ListNode * L2)
     {
         return (!L1 && !L2) ? 0 : (L1->info > L2->info) ? 1 : (L1->info < L2->info) ? -1 : compare(L1->next, L2->next) ;
     }
     static int length(ListNode * L) // O(N) so call rarely
     {
         return L == nullptr ? 0 : 1 + length(L->next);
     }
     };

     ListNode * head; // no other data members!! ­ especially no len!
};

Answer 1

首先，您对 stringToList 的实现需要稍作改动。

static ListNode * stringToList(const char *s)
{
    // What happens when the `s` is an empty string?
    // You end up storing the null character.
    ListNode * h = new ListNode(s[0], NULL);
    ListNode * c = h;

    // And then, you add s[0] again to the linked list.
    for(int i = 0; s[i] != '\0'; ++i)
    {
        c->next = new ListNode(s[i], NULL);
        c = c->next;
    }
    return h;
}

将其更改为:

static ListNode * stringToList(const char *s)
{
   ListNode * h = NULL;
   if ( s[0] != '\0' )
   {
      h = new ListNode(s[i], NULL);
      ListNode * c = h;

      // Use 1 as the starting index for the iteration
      for(int i = 1; s[i] != '\0'; ++i)
      {
         c->next = new ListNode(s[i], NULL);
         c = c->next;
      }
   }
   return h;
}

But in the case of a string object,

String::String( const String & s )

how can I possibly do that?

它与其他构造函数没有太大区别。

String::String( const String & s ) : head(NULL)
{
   ListNode* hr = s.head;
   if ( hr != NULL )
   {
      head = new ListNode(hr->info, NULL);
      List* cl = head;
      ListNode* cr = hr->next;

      for( ; cr != NULL; cr = cr->next)
      {
         cl->next = new ListNode(cr->info, NULL);
         cl = cl->next;
      }
   }
}