單鏈表C++的快速選擇算法
[英]Quickselect algorithm for singly linked list C++
問題描述
我需要一種算法，它可以在線性時間復雜度 O(n) 和恆定空間復雜度 O(1) 中找到單鏈表的中值。
編輯：單向鏈表是一個 C 風格的單向鏈表。 不允許使用 stl（沒有容器，沒有函數，禁止所有 stl，例如沒有 std::forward_list）。 不允許移動任何其他容器（如數組）中的數字。 具有 O(logn) 的空間復雜度是可以接受的，因為對於我的列表，這實際上甚至低於 100。 我也不允許使用像 nth_element 這樣的 STL 函數
基本上我有 3 * 10^6 個元素的鏈表，我需要在 3 秒內獲得中位數，所以我不能使用排序算法來對列表進行排序（這將是 O(nlogn) 並且會采取類似可能是 10-14 秒）。
我做了一些網上搜索，我發現，它的posibile找到一個std在O（n）和澳中位數::向量（1）quickselect空間compleity（最壞的情況是在為O（n ^ 2 )，但很少見)，例如： https : //www.geeksforgeeks.org/quickselect-a-simple-iterative-implementation/
但是我找不到任何為鏈表執行此操作的算法。 問題是我可以使用數組索引隨機訪問向量如果我想修改該算法，復雜性會更大，因為。 例如，當我將 pivotindex 更改為左側時，我實際上需要遍歷列表以獲取該新元素並繼續前進（這將使我的列表至少獲得 O(kn) 和一個大 k，甚至接近 O(n^ 2）……）。
編輯2：
我知道我有太多的變量，但我一直在測試不同的東西，我仍在處理我的代碼......我當前的代碼：
#include <bits/stdc++.h>
using namespace std;
template <class T> class Node {
public:
T data;
Node<T> *next;
};
template <class T> class List {
public:
Node<T> *first;
};
template <class T> T getMedianValue(List<T> & l) {
Node<T> *crt,*pivot,*incpivot;
int left, right, lung, idx, lungrel,lungrel2, left2, right2, aux, offset;
pivot = l.first;
crt = pivot->next;
lung = 1;
//lung is the lenght of the linked list (yeah it's lenght in romanian...)
//lungrel and lungrel2 are the relative lenghts of the part of
//the list I am processing, e.g: 2 3 4 in a list with 1 2 3 4 5
right = left = 0;
while (crt != NULL) {
if(crt->data < pivot->data){
aux = pivot->data;
pivot->data = crt->data;
crt->data = pivot->next->data;
pivot->next->data = aux;
pivot = pivot->next;
left++;
}
else right++;
// cout<<crt->data<<endl;
crt = crt->next;
lung++;
}
if(right > left) offset = left;
// cout<<endl;
// cout<<pivot->data<<" "<<left<<" "<<right<<endl;
// printList(l);
// cout<<endl;
lungrel = lung;
incpivot = l.first;
// offset = 0;
while(left != right){
//cout<<"parcurgere"<<endl;
if(left > right){
//cout<<endl;
//printList(l);
//cout<<endl;
//cout<<"testleft "<<incpivot->data<<" "<<left<<" "<<right<<endl;
crt = incpivot->next;
pivot = incpivot;
idx = offset;left2 = right2 = lungrel = 0;
//cout<<idx<<endl;
while(idx < left && crt!=NULL){
if(pivot->data > crt->data){
// cout<<"1crt "<<crt->data<<endl;
aux = pivot->data;
pivot->data = crt->data;
crt->data = pivot->next->data;
pivot->next->data = aux;
pivot = pivot->next;
left2++;lungrel++;
}
else {
right2++;lungrel++;
//cout<<crt->data<<" "<<right2<<endl;
}
//cout<<crt->data<<endl;
crt = crt->next;
idx++;
}
left = left2 + offset;
right = lung - left - 1;
if(right > left) offset = left;
//if(pivot->data == 18) return 18;
//cout<<endl;
//cout<<"l "<<pivot->data<<" "<<left<<" "<<right<<" "<<right2<<endl;
// printList(l);
}
else if(left < right && pivot->next!=NULL){
idx = left;left2 = right2 = 0;
incpivot = pivot->next;offset++;left++;
//cout<<endl;
//printList(l);
//cout<<endl;
//cout<<"testright "<<incpivot->data<<" "<<left<<" "<<right<<endl;
pivot = pivot->next;
crt = pivot->next;
lungrel2 = lungrel;
lungrel = 0;
// cout<<"p right"<<pivot->data<<" "<<left<<" "<<right<<endl;
while((idx < lungrel2 + offset - 1) && crt!=NULL){
if(crt->data < pivot->data){
// cout<<"crt "<<crt->data<<endl;
aux = pivot->data;
pivot->data = crt->data;
crt->data = (pivot->next)->data;
(pivot->next)->data = aux;
pivot = pivot->next;
// cout<<"crt2 "<<crt->data<<endl;
left2++;lungrel++;
}
else right2++;lungrel++;
//cout<<crt->data<<endl;
crt = crt->next;
idx++;
}
left = left2 + left;
right = lung - left - 1;
if(right > left) offset = left;
// cout<<"r "<<pivot->data<<" "<<left<<" "<<right<<endl;
// printList(l);
}
else{
//cout<<cmx<<endl;
return pivot->data;
}
}
//cout<<cmx<<endl;
return pivot->data;
}
template <class T> void printList(List<T> const & l) {
Node<T> *tmp;
if(l.first != NULL){
tmp = l.first;
while(tmp != NULL){
cout<<tmp->data<<" ";
tmp = tmp->next;
}
}
}
template <class T> void push_front(List<T> & l, int x)
{
Node<T>* tmp = new Node<T>;
tmp->data = x;
tmp->next = l.first;
l.first = tmp;
}
int main(){
List<int> l;
int n = 0;
push_front(l, 19);
push_front(l, 12);
push_front(l, 11);
push_front(l, 101);
push_front(l, 91);
push_front(l, 21);
push_front(l, 9);
push_front(l, 6);
push_front(l, 25);
push_front(l, 4);
push_front(l, 18);
push_front(l, 2);
push_front(l, 8);
push_front(l, 10);
push_front(l, 200);
push_front(l, 225);
push_front(l, 170);
printList(l);
n=getMedianValue(l);
cout<<endl;
cout<<n;
return 0;
}
您對如何將快速選擇調整到單獨列出的鏈接或其他適用於我的問題的算法有任何建議嗎？
2 個解決方案
解決方案1
3 已采納 2020-03-28 23:23:45
在您的問題中，您提到您在選擇不在列表開頭的樞軸時遇到問題，因為這需要遍歷列表。 如果你做對了，你只需要遍歷整個列表兩次：
- 一次用於查找列表的中間和末尾以選擇一個好的樞軸（例如使用“三的中位數”規則）
- 一次用於實際排序
如果您不太關心選擇一個好的樞軸，並且您很高興只需選擇列表的第一個元素作為樞軸（這會導致最壞的情況下 O(n^2)時間復雜度，如果數據已經排序）。
如果您通過維護指向末尾的指針來記住第一次遍歷列表的末尾，那么您永遠不必再次遍歷它來找到末尾。 此外，如果您使用的是標准Lomuto 分區方案（我沒有使用，原因如下所述），那么您還必須維護指向列表的兩個指針，它們表示標准 Lomuto 分區方案的i和j索引。 通過使用這些指針，永遠不必遍歷列表來訪問單個元素。
此外，如果您維護一個指向每個分區的中間和末尾的指針，那么當您稍后必須對這些分區之一進行排序時，您將不必再次遍歷該分區以找到中間和末尾。
我現在已經為鏈表創建了我自己的QuickSelect算法實現，我已經發布在下面。
既然你說鏈表是單向鏈表，不能升級為雙向鏈表，我不能使用 霍爾分區方案，因為向后迭代單向鏈表非常昂貴。 因此，我改用效率通常較低的Lomuto 分區方案。
使用 Lomuto 分區方案時，通常選擇第一個元素或最后一個元素作為主元。 但是，選擇其中任何一個都有缺點，即排序后的數據將導致算法的最壞情況時間復雜度為 O(n^2)。 這可以通過根據“三的中位數”規則選擇一個主元來防止，即從第一個元素、中間元素和最后一個元素的中值中選擇一個主元。 因此，在我的實現中，我使用了這個“三的中位數”規則。
此外，Lomuto 分區方案通常會創建兩個分區，一個用於小於樞軸的值，另一個用於大於或等於樞軸的值。 但是，如果所有值都相同，這將導致 O(n^2) 的最壞情況時間復雜度。 因此，在我的實現中，我創建了三個分區，一個用於小於樞軸的值，一個用於大於樞軸的值，另一個用於等於樞軸的值。
盡管這些措施並沒有完全消除 O(n^2) 的最壞情況時間復雜度的可能性，但它們至少使它極不可能發生（除非輸入是由惡意攻擊者提供的）。 為了保證 O(n) 的時間復雜度，必須使用更復雜的主元選擇算法，例如中值的中位數。
我遇到的一個重要問題是，對於偶數個元素，中位數定義為兩個“中間”或“中位數”元素的算術平均值。 出於這個原因，我不能簡單地編寫類似於std::nth_element的函數，因為例如，如果元素總數為 14，那么我將尋找第 7 和第 8 大元素。 這意味着我必須兩次調用這樣的函數，這將是低效的。 因此，我編寫了一個可以同時搜索兩個“中值”元素的函數。 盡管這使代碼更加復雜，但與不必兩次調用相同函數的優勢相比，由於額外的代碼復雜性而導致的性能損失應該是最小的。
請注意，盡管我的實現在 C++ 編譯器上完美編譯，但我不會稱其為教科書 C++ 代碼，因為問題指出我不允許使用 C++ 標准模板庫中的任何內容。 因此，我的代碼是 C 代碼和 C++ 代碼的混合體。
在下面的代碼中，我只使用標准模板庫（特別是函數std::nth_element ）來測試我的算法和驗證結果。 我在我的實際算法中沒有使用這些函數中的任何一個。
#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;
}
我已經成功地用一百萬個隨機生成的元素測試了我的代碼，它幾乎立即找到了正確的中位數。
解決方案2
1 2020-03-20 20:31:35
所以你可以做的是使用迭代器來保持位置。 我已經編寫了上面的算法來處理 std::forward_list。 我知道這並不完美，但很快就把它寫下來並希望它有所幫助。
int partition(int leftPos, int rightPos, std::forward_list<int>::iterator& currIter,
std::forward_list<int>::iterator lowIter, std::forward_list<int>::iterator highIter) {
auto iter = lowIter;
int i = leftPos - 1;
for(int j = leftPos; j < rightPos - 1; j++) {
if(*iter <= *highIter) {
++currIter;
++i;
std::iter_swap(currIter, iter);
}
iter++;
}
std::forward_list<int>::iterator newIter = currIter;
std::iter_swap(++newIter, highIter);
return i + 1;
}
std::forward_list<int>::iterator kthSmallest(std::forward_list<int>& list,
std::forward_list<int>::iterator left, std::forward_list<int>::iterator right, int size, int k) {
int leftPos {0};
int rightPos {size};
int pivotPos {0};
std::forward_list<int>::iterator resetIter = left;
std::forward_list<int>::iterator currIter = left;
++left;
while(leftPos <= rightPos) {
pivotPos = partition(leftPos, rightPos, currIter, left, right);
if(pivotPos == (k-1)) {
return currIter;
} else if(pivotPos > (k-1)) {
right = currIter;
rightPos = pivotPos - 1;
} else {
left = currIter;
++left;
resetIter = left;
++left;
leftPos = pivotPos + 1;
}
currIter = resetIter;
}
return list.end();
}
調用第 k 個迭代器時，左迭代器應該比您打算開始的位置少 1。 這使我們能夠在partition()落后於low一位。 下面是一個執行它的例子：
int main() {
std::forward_list<int> list {10, 12, 12, 13, 4, 5, 8, 11, 6, 26, 15, 21};
auto startIter = list.before_begin();
int k = 6;
int size = getSize(list);
auto kthIter = kthSmallest(list, startIter, getEnd(list), size - 1, k);
std::cout << k << "th smallest: " << *kthIter << std::endl;
return 0;
}
第六小：10
單鏈表 C++
聲明:本站的技術帖子網頁，遵循CC BY-SA 4.0協議，如果您需要轉載，請注明本站網址或者原文地址。任何問題請咨詢:yoyou2525@163.com.