在您的问题中，您提到您无法选择不在列表开头的枢轴，因为这需要遍历列表。如果你做对了，你只需要遍历整个列表两次：
- 一次用于查找列表的中间和末尾以选择一个好的枢轴（例如，使用"median-of-three" 规则）
- 实际排序一次
如果您不太关心选择一个好的枢轴并且您很高兴只需选择列表的第一个元素作为枢轴（这会导致最坏的情况 O(n^2) @987654322 @如果数据已经排序）。
如果您通过维护指向末尾的指针第一次遍历它时记住了列表的末尾，那么您永远不必再次遍历它以找到末尾。此外，如果您使用标准的 Lomuto partition scheme（我没有使用它，原因如下），那么您还必须维护两个指向列表的指针，它们代表标准 Lomuto 分区的 i 和 j 索引方案。通过使用这些指针，永远不必遍历列表来访问单个元素。
此外，如果您维护一个指向每个分区的中间和结尾的指针，那么，当您以后必须对这些分区中的一个进行排序时，您将不必再次遍历该分区来找到中间和结尾。
我现在已经为链表创建了自己的QuickSelect 算法实现，我在下面发布了该算法。
既然你说链表是单链表并且不能升级为双链表，我不能使用Hoare partition scheme，因为向后迭代单链表非常昂贵。因此，我改用通常效率较低的Lomuto partition scheme。
使用 Lomuto 分区方案时，通常选择第一个元素或最后一个元素作为枢轴。但是，选择其中任何一个都有一个缺点，即排序的数据将导致算法具有 O(n^2) 的最坏情况时间复杂度。这可以通过根据"median-of-three" rule选择一个pivot来防止，即从第一个元素、中间元素和最后一个元素的中间值中选择一个pivot。因此，在我的实现中，我使用的是“三中位数”规则。
此外，Lomuto 分区方案通常会创建两个分区，一个用于小于基准的值，另一个用于大于或等于基准的值。但是，如果所有值都相同，这将导致 O(n^2) 的最坏情况时间复杂度。因此，在我的实现中，我创建了三个分区，一个用于小于枢轴的值，一个用于大于枢轴的值，一个用于等于枢轴的值。
虽然这些措施并不能完全消除 O(n^2) 的最坏情况时间复杂度的可能性，但它们至少使这种可能性极小（除非输入是由恶意攻击者提供的）。为了保证 O(n) 的时间复杂度，必须使用更复杂的枢轴选择算法，例如median of medians。
我遇到的一个重要问题是，对于偶数个元素，median 被定义为两个“中间”或“中间”元素的arithmetic mean。出于这个原因，我不能简单地编写类似于std::nth_element 的函数，因为例如，如果元素总数为 14，那么我将寻找第 7 和第 8 大的元素。这意味着我必须调用这样的函数两次，这将是低效的。因此，我编写了一个可以同时搜索两个“中值”元素的函数。虽然这使代码更加复杂，但与不必调用相同函数两次的优势相比，由于额外的代码复杂性而导致的性能损失应该是最小的。
请注意，尽管我的实现可以在 C++ 编译器上完美编译，但我不会将其称为教科书 C++ 代码，因为问题表明我不允许使用 C++ 标准模板库中的任何内容。因此，我的代码是 C 代码和 C++ 代码的混合体。
在下面的代码中，我只使用标准模板库（特别是函数std::nth_element）来测试我的算法并验证结果。在我的实际算法中，我没有使用任何这些函数。
#include <iostream>
#include <iomanip>
#include <cassert>
// The following two headers are only required for testing the algorithm and verifying
// the correctness of its results. They are not used in the algorithm itself.
#include <random>
#include <algorithm>
// The following setting can be changed to print extra debugging information
// possible settings:
// 0: no extra debugging information
// 1: print the state and length of all partitions in every loop iteraton
// 2: additionally print the contents of all partitions (if they are not too big)
#define PRINT_DEBUG_LEVEL 0
template <typename T>
struct Node
{
T data;
Node<T> *next;
};
// NOTE:
// The return type is not necessarily the same as the data type. The reason for this is
// that, for example, the data type "int" requires a "double" as a return type, so that
// the arithmetic mean of "3" and "6" returns "4.5".
// This function may require template specializations to handle overflow or wrapping.
template<typename T, typename U>
U arithmetic_mean( const T &first, const T &second )
{
return ( static_cast<U>(first) + static_cast<U>(second) ) / 2;
}
//the main loop of the function find_median can be in one of the following three states
enum LoopState
{
//we are looking for one median value
LOOPSTATE_LOOKINGFORONE,
//we are looking for two median values, and the returned median
//will be the arithmetic mean of the two
LOOPSTATE_LOOKINGFORTWO,
//one of the median values has been found, but we are still searching for
//the second one
LOOPSTATE_FOUNDONE
};
template <
typename T, //type of the data
typename U //type of the return value
>
U find_median( Node<T> *list )
{
//This variable points to the pointer to the first element of the current partition.
//During the partition phase, the linked list will be broken and reassembled afterwards, so
//the pointer this pointer points to will be nullptr until it is reassembled.
Node<T> **pp_start = &list;
//This pointer represents nothing more than the cached value of *pp_start and it is
//not always valid
Node<T> *p_start = *pp_start;
//These pointers are maintained for accessing the middle of the list for selecting a pivot
//using the "median-of-three" rule.
Node<T> *p_middle;
Node<T> *p_end;
//result is not defined if list is empty
assert( p_start != nullptr );
//in the main loop, this variable always holds the number of elements in the current partition
int num_total = 1;
// First, we must traverse the entire linked list in order to determine the number of elements,
// in order to calculate k1 and k2. If it is odd, then the median is defined as the k'th smallest
// element where k = n / 2. If the number of elements is even, then the median is defined as the
// arithmetic mean of the k'th element and the (k+1)'th element.
// We also set a pointer to the nodes in the middle and at the end, which will be required later
// for selecting a pivot according to the "median-of-three" rule.
p_middle = p_start;
for ( p_end = p_start; p_end->next != nullptr; p_end = p_end->next )
{
num_total++;
if ( num_total % 2 == 0 ) p_middle = p_middle->next;
}
// find out whether we are looking for only one or two median values
enum LoopState loop_state = num_total % 2 == 0 ? LOOPSTATE_LOOKINGFORTWO : LOOPSTATE_LOOKINGFORONE;
//set k to the index of the middle element, or if there are two middle elements, to the left one
int k = ( num_total - 1 ) / 2;
// If we are looking for two median values, but we have only found one, then this variable will
// hold the value of the one we found. Whether we have found one can be determined by the state of
// the variable loop_state.
T val_found;
for (;;)
{
//make p_start cache the value of *pp_start again, because a previous iteration of the loop
//may have changed the value of pp_start
p_start = *pp_start;
assert( p_start != nullptr );
assert( p_middle != nullptr );
assert( p_end != nullptr );
assert( num_total != 0 );
if ( num_total == 1 )
{
switch ( loop_state )
{
case LOOPSTATE_LOOKINGFORONE:
return p_start->data;
case LOOPSTATE_FOUNDONE:
return arithmetic_mean<T,U>( val_found, p_start->data );
default:
assert( false ); //this should be unreachable
}
}
//select the pivot according to the "median-of-three" rule
T pivot;
if ( p_start->data < p_middle->data )
{
if ( p_middle->data < p_end->data )
pivot = p_middle->data;
else if ( p_start->data < p_end->data )
pivot = p_end->data;
else
pivot = p_start->data;
}
else
{
if ( p_start->data < p_end->data )
pivot = p_start->data;
else if ( p_middle->data < p_end->data )
pivot = p_end->data;
else
pivot = p_middle->data;
}
#if PRINT_DEBUG_LEVEL >= 1
//this line is conditionally compiled for extra debugging information
std::cout << "\nmedian of three: " << (*pp_start)->data << " " << p_middle->data << " " << p_end->data << " ->" << pivot << std::endl;
#endif
// We will be dividing the current partition into 3 new partitions (less-than,
// equal-to and greater-than) each represented as a linked list. Each list
// requires a pointer to the start of the list and a pointer to the pointer at
// the end of the list to write the address of new elements to. Also, when
// traversing the lists, we need to keep a pointer to the middle of the list,
// as this information will be required for selecting a new pivot in the next
// iteration of the loop. The latter is not required for the equal-to partition,
// as it would never be used.
Node<T> *p_less = nullptr, **pp_less_end = &p_less, **pp_less_middle = &p_less;
Node<T> *p_equal = nullptr, **pp_equal_end = &p_equal;
Node<T> *p_greater = nullptr, **pp_greater_end = &p_greater, **pp_greater_middle = &p_greater;
// These pointers are only used as a cache to the location of the end node.
// Despite their similar name, their function is quite different to pp_less_end
// and pp_greater_end.
Node<T> *p_less_end = nullptr;
Node<T> *p_greater_end = nullptr;
// counter for the number of elements in each partition
int num_less = 0;
int num_equal = 0;
int num_greater = 0;
// NOTE:
// The following loop will temporarily split the linked list. It will be merged later.
Node<T> *p_next_node = p_start;
//the following line isn't necessary; it is only used to clarify that the pointers no
//longer point to anything meaningful
*pp_start = p_start = nullptr;
for ( int i = 0; i < num_total; i++ )
{
assert( p_next_node != nullptr );
Node<T> *p_current_node = p_next_node;
p_next_node = p_next_node->next;
if ( p_current_node->data < pivot )
{
//link node to pp_less
assert( *pp_less_end == nullptr );
*pp_less_end = p_less_end = p_current_node;
pp_less_end = &p_current_node->next;
p_current_node->next = nullptr;
num_less++;
if ( num_less % 2 == 0 )
{
pp_less_middle = &(*pp_less_middle)->next;
}
}
else if ( p_current_node->data == pivot )
{
//link node to pp_equal
assert( *pp_equal_end == nullptr );
*pp_equal_end = p_current_node;
pp_equal_end = &p_current_node->next;
p_current_node->next = nullptr;
num_equal++;
}
else
{
//link node to pp_greater
assert( *pp_greater_end == nullptr );
*pp_greater_end = p_greater_end = p_current_node;
pp_greater_end = &p_current_node->next;
p_current_node->next = nullptr;
num_greater++;
if ( num_greater % 2 == 0 )
{
pp_greater_middle = &(*pp_greater_middle)->next;
}
}
}
assert( num_total == num_less + num_equal + num_greater );
assert( num_equal >= 1 );
#if PRINT_DEBUG_LEVEL >= 1
//this section is conditionally compiled for extra debugging information
{
std::cout << std::setfill( '0' );
switch ( loop_state )
{
case LOOPSTATE_LOOKINGFORONE:
std::cout << "LOOPSTATE_LOOKINGFORONE k = " << k << "\n";
break;
case LOOPSTATE_LOOKINGFORTWO:
std::cout << "LOOPSTATE_LOOKINGFORTWO k = " << k << "\n";
break;
case LOOPSTATE_FOUNDONE:
std::cout << "LOOPSTATE_FOUNDONE k = " << k << " val_found = " << val_found << "\n";
}
std::cout << "partition lengths: ";
std::cout <<
std::setw( 2 ) << num_less << " " <<
std::setw( 2 ) << num_equal << " " <<
std::setw( 2 ) << num_greater << " " <<
std::setw( 2 ) << num_total << "\n";
#if PRINT_DEBUG_LEVEL >= 2
Node<T> *p;
std::cout << "less: ";
if ( num_less > 10 )
std::cout << "too many to print";
else
for ( p = p_less; p != nullptr; p = p->next ) std::cout << p->data << " ";
std::cout << "\nequal: ";
if ( num_equal > 10 )
std::cout << "too many to print";
else
for ( p = p_equal; p != nullptr; p = p->next ) std::cout << p->data << " ";
std::cout << "\ngreater: ";
if ( num_greater > 10 )
std::cout << "too many to print";
else
for ( p = p_greater; p != nullptr; p = p->next ) std::cout << p->data << " ";
std::cout << "\n\n" << std::flush;
#endif
std::cout << std::flush;
}
#endif
//insert less-than partition into list
assert( *pp_start == nullptr );
*pp_start = p_less;
//insert equal-to partition into list
assert( *pp_less_end == nullptr );
*pp_less_end = p_equal;
//insert greater-than partition into list
assert( *pp_equal_end == nullptr );
*pp_equal_end = p_greater;
//link list to previously cut off part
assert( *pp_greater_end == nullptr );
*pp_greater_end = p_next_node;
//if less-than partition is large enough to hold both possible median values
if ( k + 2 <= num_less )
{
//set the next iteration of the loop to process the less-than partition
//pp_start is already set to the desired value
p_middle = *pp_less_middle;
p_end = p_less_end;
num_total = num_less;
}
//else if less-than partition holds one of both possible median values
else if ( k + 1 == num_less )
{
if ( loop_state == LOOPSTATE_LOOKINGFORTWO )
{
//the equal_to partition never needs sorting, because all members are already equal
val_found = p_equal->data;
loop_state = LOOPSTATE_FOUNDONE;
}
//set the next iteration of the loop to process the less-than partition
//pp_start is already set to the desired value
p_middle = *pp_less_middle;
p_end = p_less_end;
num_total = num_less;
}
//else if equal-to partition holds both possible median values
else if ( k + 2 <= num_less + num_equal )
{
//the equal_to partition never needs sorting, because all members are already equal
if ( loop_state == LOOPSTATE_FOUNDONE )
return arithmetic_mean<T,U>( val_found, p_equal->data );
return p_equal->data;
}
//else if equal-to partition holds one of both possible median values
else if ( k + 1 == num_less + num_equal )
{
switch ( loop_state )
{
case LOOPSTATE_LOOKINGFORONE:
return p_equal->data;
case LOOPSTATE_LOOKINGFORTWO:
val_found = p_equal->data;
loop_state = LOOPSTATE_FOUNDONE;
k = 0;
//set the next iteration of the loop to process the greater-than partition
pp_start = pp_equal_end;
p_middle = *pp_greater_middle;
p_end = p_greater_end;
num_total = num_greater;
break;
case LOOPSTATE_FOUNDONE:
return arithmetic_mean<T,U>( val_found, p_equal->data );
}
}
//else both possible median values must be in the greater-than partition
else
{
k = k - num_less - num_equal;
//set the next iteration of the loop to process the greater-than partition
pp_start = pp_equal_end;
p_middle = *pp_greater_middle;
p_end = p_greater_end;
num_total = num_greater;
}
}
}
// NOTE:
// The following code is not part of the algorithm, but is only intended to test the algorithm
// This simple class is designed to contain a singly-linked list
template <typename T>
class List
{
public:
List() : first( nullptr ) {}
// the following is required to abide by the rule of three/five/zero
// see: https://en.cppreference.com/w/cpp/language/rule_of_three
List( const List<T> & ) = delete;
List( const List<T> && ) = delete;
List<T>& operator=( List<T> & ) = delete;
List<T>& operator=( List<T> && ) = delete;
~List()
{
Node<T> *p = first;
while ( p != nullptr )
{
Node<T> *temp = p;
p = p->next;
delete temp;
}
}
void push_front( int data )
{
Node<T> *temp = new Node<T>;
temp->data = data;
temp->next = first;
first = temp;
}
//member variables
Node<T> *first;
};
int main()
{
//generated random numbers will be between 0 and 2 billion (fits in 32-bit signed int)
constexpr int min_val = 0;
constexpr int max_val = 2*1000*1000*1000;
//will allocate array for 1 million ints and fill with random numbers
constexpr int num_values = 1*1000*1000;
//this class contains the singly-linked list and is empty for now
List<int> l;
double result;
//These variables are used for random number generation
std::random_device rd;
std::mt19937 gen( rd() );
std::uniform_int_distribution<> dis( min_val, max_val );
try
{
//fill array with random data
std::cout << "Filling array with random data..." << std::flush;
auto unsorted_data = std::make_unique<int[]>( num_values );
for ( int i = 0; i < num_values; i++ ) unsorted_data[i] = dis( gen );
//fill the singly-linked list
std::cout << "done\nFilling linked list..." << std::flush;
for ( int i = 0; i < num_values; i++ ) l.push_front( unsorted_data[i] );
std::cout << "done\nCalculating median using STL function..." << std::flush;
//calculate the median using the functions provided by the C++ standard template library.
//Note: this is only done to compare the results with the algorithm provided in this file
if ( num_values % 2 == 0 )
{
int median1, median2;
std::nth_element( &unsorted_data[0], &unsorted_data[(num_values - 1) / 2], &unsorted_data[num_values] );
median1 = unsorted_data[(num_values - 1) / 2];
std::nth_element( &unsorted_data[0], &unsorted_data[(num_values - 0) / 2], &unsorted_data[num_values] );
median2 = unsorted_data[(num_values - 0) / 2];
result = arithmetic_mean<int,double>( median1, median2 );
}
else
{
int median;
std::nth_element( &unsorted_data[0], &unsorted_data[(num_values - 0) / 2], &unsorted_data[num_values] );
median = unsorted_data[(num_values - 0) / 2];
result = static_cast<int>(median);
}
std::cout << "done\nMedian according to STL function: " << std::setprecision( 12 ) << result << std::endl;
// NOTE: Since the STL functions only sorted the array, but not the linked list, the
// order of the linked list is still random and not pre-sorted.
//calculate the median using the algorithm provided in this file
std::cout << "Starting algorithm" << std::endl;
result = find_median<int,double>( l.first );
std::cout << "The calculated median is: " << std::setprecision( 12 ) << result << std::endl;
std::cout << "Cleaning up\n\n" << std::flush;
}
catch ( std::bad_alloc )
{
std::cerr << "Error: Unable to allocate sufficient memory!" << std::endl;
return -1;
}
return 0;
}
我已经用一百万个随机生成的元素成功地测试了我的代码，它几乎立即找到了正确的中位数。