【发布时间】：2021-06-13 06:47:57
【问题描述】：
这是一个类的作业，对于这个程序，我需要实现链表的二进制搜索，我该怎么做？是否可以确定和访问列表的中间元素，并仅关注左侧或 列表的右半部分？ 如何实现链表的二分查找？
搜索.h
#ifndef SEARCHING_H_
#define SEARCHING_H_
#include "Vector.h"
#include "List.h"
using namespace std;
template <typename T>
void print_vector(const Vector<T>& vec)
{
for (int i = 0; i < vec.size(); i++)
cout << vec[i] << " ";
cout << endl;
return;
}
// LINEAR SEARCH OF VECTOR
// return index at which target found, if not found return -1;
template <typename T>
int linear_search_V(const Vector<T>& vec, const T& target, int& ops)
{
ops = 0;
for (int i = 0; i < vec.size(); i++) {
ops++;
if (vec[i] == target)
return i;
}
return -1;
}
// LINEAR SEARCH OF LINKED LIST
// return iterator to node at which target
// found in lst; else return iterator at end() of lst;
template <typename T>
typename List<T>::const_iterator linear_search_L(const List<T>& lst, const T& target, int& ops)
{
ops = 0;
typename List<T>::const_iterator itr;
for (itr = lst.begin(); itr != lst.end(); ++itr) {
ops++;
if (*itr++ == target)
return itr;
}
return lst.end();
}
// BINARY SEARCH OF VECTOR;
// returns index at which target found, else -1;
template <typename T>
int binary_search_V(const Vector<T>& vec, const T& target, int& ops)
{
ops = 0;
int low = 0;
int high = vec.size() - 1;
while (low <= high) {
ops++;
int mid = (low + high) / 2;
if (vec[mid] < target)
low = mid + 1;
else if (vec[mid] > target)
high = mid - 1;
else
return mid;
}
return -1;
}
// BINARY SEARCH OF LINKED LIST
// returns iterator at target value found; iterator end() else;
template <typename T>
typename List<T>::const_iterator binary_search_L(const List<T> lst, const T& target, int& ops)
{
ops = 0;
//implement function for binary search in Linked List:
}
#endif
//this is the List.h i need to use for the binary search in a linked list
#ifndef LIST_H
#define LIST_H
//#include <algorithm>
using namespace std;
template <typename T>
class List
{
private:
// The basic doubly linked list node.
// Nested inside of List, can be public
// because the Node is itself private
struct Node
{
T data;
Node *prev;
Node *next;
Node( const T & d = T{ }, Node * p = nullptr, Node * n = nullptr )
: data{ d }, prev{ p }, next{ n } { }
Node( T && d, Node * p = nullptr, Node * n = nullptr )
: data{ std::move( d ) }, prev{ p }, next{ n } { }
};
public:
class const_iterator
{
public:
// Public constructor for const_iterator.
const_iterator( ) : current{ nullptr }
{ }
// Return the T stored at the current position.
// For const_iterator, this is an accessor with a
// const reference return type.
const T & operator* ( ) const
{ return retrieve( ); }
const_iterator & operator++ ( )
{
current = current->next;
return *this;
}
const_iterator operator++ ( int )
{
const_iterator old = *this;
++( *this );
return old;
}
const_iterator & operator-- ( )
{
current = current->prev;
return *this;
}
const_iterator operator-- ( int )
{
const_iterator old = *this;
--( *this );
return old;
}
bool operator== ( const const_iterator & rhs ) const
{ return current == rhs.current; }
bool operator!= ( const const_iterator & rhs ) const
{ return !( *this == rhs ); }
// iterators side-by-side
bool operator > (const const_iterator & rhs) const
{
return current->prev == rhs.current;
}
protected:
Node *current;
// Protected helper in const_iterator that returns the T
// stored at the current position. Can be called by all
// three versions of operator* without any type conversions.
T & retrieve( ) const
{ return current->data; }
// Protected constructor for const_iterator.
// Expects a pointer that represents the current position.
const_iterator( Node *p ) : current{ p }
{ }
friend class List<T>;
};
class iterator : public const_iterator
{
public:
// Public constructor for iterator.
// Calls the base-class constructor.
// Must be provided because the private constructor
// is written; otherwise zero-parameter constructor
// would be disabled.
iterator( )
{ }
T & operator* ( )
{ return const_iterator::retrieve( ); }
.
const T & operator* ( ) const
{ return const_iterator::operator*( ); }
iterator & operator++ ( )
{
this->current = this->current->next;
return *this;
}
iterator operator++ ( int )
{
iterator old = *this;
++( *this );
return old;
}
iterator & operator-- ( )
{
this->current = this->current->prev;
return *this;
}
iterator operator-- ( int )
{
iterator old = *this;
--( *this );
return old;
}
protected:
// Protected constructor for iterator.
// Expects the current position.
iterator( Node *p ) : const_iterator{ p }
{ }
friend class List<T>;
};
public:
List( )
{ init( ); }
~List( )
{
clear( );
delete head;
delete tail;
}
List( const List & rhs )
{
init( );
for( auto & x : rhs )
push_back( x );
}
List & operator= ( const List & rhs )
{
List copy = rhs;
std::swap( *this, copy );
return *this;
}
// keep for compiler
List(List&& rhs)
: theSize{ rhs.theSize }, head{ rhs.head }, tail{ rhs.tail }
{
rhs.theSize = 0;
rhs.head = nullptr;
rhs.tail = nullptr;
}
// keep for compiler
List& operator= (List&& rhs)
{
std::swap(theSize, rhs.theSize);
std::swap(head, rhs.head);
std::swap(tail, rhs.tail);
return *this;
}
// Return iterator representing beginning of list.
// Mutator version is first, then accessor version.
iterator begin( )
{ return iterator( head->next ); }
const_iterator begin( ) const
{ return const_iterator( head->next ); }
// Return iterator representing endmarker of list.
// Mutator version is first, then accessor version.
iterator end( )
{ return iterator( tail ); }
const_iterator end( ) const
{ return const_iterator( tail ); }
// Return number of elements currently in the list.
int size( ) const
{ return theSize; }
// Return true if the list is empty, false otherwise.
bool empty( ) const
{ return size( ) == 0; }
void clear( )
{
while( !empty( ) )
pop_front( );
}
// front, back, push_front, push_back, pop_front, and pop_back
// are the basic double-ended queue operations.
T & front( )
{ return *begin( ); }
const T & front( ) const
{ return *begin( ); }
T & back( )
{ return *--end( ); }
const T & back( ) const
{ return *--end( ); }
void push_front( const T & x )
{ insert( begin( ), x ); }
void push_back( const T & x )
{ insert( end( ), x ); }
void pop_front( )
{ erase( begin( ) ); }
void pop_back( )
{ erase( --end( ) ); }
// Insert x before itr.
iterator insert( iterator itr, const T & x )
{
Node *p = itr.current;
++theSize;
return iterator( p->prev = p->prev->next = new Node{ x, p->prev, p } );
}
// Erase item at itr.
iterator erase( iterator itr )
{
Node *p = itr.current;
iterator retVal( p->next );
p->prev->next = p->next;
p->next->prev = p->prev;
delete p;
--theSize;
return retVal;
}
iterator erase( iterator from, iterator to )
{
for( iterator itr = from; itr != to; )
itr = erase( itr );
return to;
}
private:
int theSize;
Node *head;
Node *tail;
void init( )
{
theSize = 0;
head = new Node;
tail = new Node;
head->next = tail;
tail->prev = head;
}
};
#endif
【问题讨论】：
-
泪流满面。链接列表和排序不是理想的伴侣。
-
单链表还是双链表？
-
我建议使用节点数而不是索引。对于单个链表，如果键小于中点的值，则必须从顶部开始。如果key大于中点的值，可以继续迭代。
-
为什么要这样做？这是c ++类中的问题吗？如果是这样，目的可能是说明二分搜索与链表不兼容。如果不是，您应该根据实际要求使用不同的容器。
-
在双链表中进行二分搜索更容易，但仍然很痛苦。您只需向前或向后（从当前中点）迭代到下一个中点。这里的关键短语是“迭代到中点”。有了支持随机访问的容器，就不需要迭代了。
标签： c++ computer-science