鏈表 - AppendLastNToFirst (C++)，我只寫了問題的邏輯。 請告訴我更正

Question

給定一個鏈表和一個整數 n，將 LL 的最后 n 個元素追加到前面。 索引從0開始，你不需要打印元素，只需更新元素並返回更新后的LL的頭部。 假設給定的 n 將小於 LL 的長度。

node* append_LinkedList(node* head,int n)
{
    int len=0;
    node* temp=head;
    while(temp->next!=NULL){
        temp=temp->next;
        len++;
    }
    for(int i=1; i<=len-n; i++){
        temp=temp->next;
    }
    node* curr=temp;
    node* curr1=curr->next;
    curr->next=NULL;
    while(curr1->next!=NULL){
        curr1=curr1->next;
    }
    curr1->next=head;
    return head;
}

Answer 1

如果我正確理解了作業，那么您需要類似於下面演示程序中所示的內容。

我使用了模板結構，但如果您願意，您可以輕松刪除模板規范的任何提及。

我將函數命名為append_n 。

這個給你。

#include <iostream>
#include <iterator>

template <typename T>
struct node
{
    T data;
    node *next;
};

template <typename T, typename InputIterator>
void assign( node<T> * &head, InputIterator first, InputIterator last )
{
    while ( head != nullptr )
    {
        node<T> *current = head;
        head = head->next;
        delete current;
    }

    for ( node<T> **current = &head; first != last; ++first )
    {
        *current = new node<T>{ *first, nullptr };
        current = &( *current )->next;
    }
}

template <typename T>
void append_n( node<T> * &head, size_t n )
{
    if ( n != 0 )
    {
        node<T> **last = &head;

        while ( *last != nullptr && n-- )
        {
            last = &( *last )->next;
        }

        if ( *last != nullptr )
        {
            node<T> **first = &head;

            while ( *last != nullptr )
            {
                first = &( *first )->next;
                last  = &( *last )->next;
            }

            *last = head;
            head = *first;
            *first = nullptr;
        }
    }       
}

template <typename T>
std::ostream & operator <<( std::ostream &os, node<T> * &head )
{
    for ( const node<T> *current = head; current != nullptr; current = current->next )
    {
        os << current->data << " -> ";
    }

    return os << "null";
}

int main() 
{
    node<int> *head = nullptr;
    int a[] = { 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 };
    const size_t N = sizeof( a ) / sizeof( *a );

    assign( head, std::begin( a ), std::end( a ) );

    std::cout << head << '\n';

    for ( size_t i = 0, n = 1; i < N; i += n )
    {
        append_n( head, n );

        std::cout << head << '\n';
    }

    std::cout << '\n';

    for ( size_t i = 0, n = 2; i < N; i += n )
    {
        append_n( head, n );

        std::cout << head << '\n';
    }

    std::cout << '\n';

    for ( size_t i = 0, n = 5; i < N; i += n )
    {
        append_n( head, n );

        std::cout << head << '\n';
    }

    std::cout << '\n';

    return 0;
}

程序輸出是

0 -> 1 -> 2 -> 3 -> 4 -> 5 -> 6 -> 7 -> 8 -> 9 -> null
9 -> 0 -> 1 -> 2 -> 3 -> 4 -> 5 -> 6 -> 7 -> 8 -> null
8 -> 9 -> 0 -> 1 -> 2 -> 3 -> 4 -> 5 -> 6 -> 7 -> null
7 -> 8 -> 9 -> 0 -> 1 -> 2 -> 3 -> 4 -> 5 -> 6 -> null
6 -> 7 -> 8 -> 9 -> 0 -> 1 -> 2 -> 3 -> 4 -> 5 -> null
5 -> 6 -> 7 -> 8 -> 9 -> 0 -> 1 -> 2 -> 3 -> 4 -> null
4 -> 5 -> 6 -> 7 -> 8 -> 9 -> 0 -> 1 -> 2 -> 3 -> null
3 -> 4 -> 5 -> 6 -> 7 -> 8 -> 9 -> 0 -> 1 -> 2 -> null
2 -> 3 -> 4 -> 5 -> 6 -> 7 -> 8 -> 9 -> 0 -> 1 -> null
1 -> 2 -> 3 -> 4 -> 5 -> 6 -> 7 -> 8 -> 9 -> 0 -> null
0 -> 1 -> 2 -> 3 -> 4 -> 5 -> 6 -> 7 -> 8 -> 9 -> null

8 -> 9 -> 0 -> 1 -> 2 -> 3 -> 4 -> 5 -> 6 -> 7 -> null
6 -> 7 -> 8 -> 9 -> 0 -> 1 -> 2 -> 3 -> 4 -> 5 -> null
4 -> 5 -> 6 -> 7 -> 8 -> 9 -> 0 -> 1 -> 2 -> 3 -> null
2 -> 3 -> 4 -> 5 -> 6 -> 7 -> 8 -> 9 -> 0 -> 1 -> null
0 -> 1 -> 2 -> 3 -> 4 -> 5 -> 6 -> 7 -> 8 -> 9 -> null

5 -> 6 -> 7 -> 8 -> 9 -> 0 -> 1 -> 2 -> 3 -> 4 -> null
0 -> 1 -> 2 -> 3 -> 4 -> 5 -> 6 -> 7 -> 8 -> 9 -> null

如果在函數返回指向頭節點的指針時使用函數聲明，則其定義可以如下面的演示程序所示。

#include <iostream>
#include <iterator>

template <typename T>
struct node
{
    T data;
    node *next;
};

template <typename T, typename InputIterator>
void assign( node<T> * &head, InputIterator first, InputIterator last )
{
    while ( head != nullptr )
    {
        node<T> *current = head;
        head = head->next;
        delete current;
    }

    for ( node<T> **current = &head; first != last; ++first )
    {
        *current = new node<T>{ *first, nullptr };
        current = &( *current )->next;
    }
}

template <typename T>
node<T> * append_n( node<T> * head, size_t n )
{
    if ( n != 0 )
    {
        node<T> *last = head;

        while ( last != nullptr && n-- )
        {
            last = last->next;
        }

        if ( last != nullptr )
        {
            node<T> *first = head;

            while ( last->next != nullptr )
            {
                first = first->next;
                last  = last->next;
            }

            last->next = head;
            head = first->next;
            first->next = nullptr;
        }
    }

    return head;
}

template <typename T>
std::ostream & operator <<( std::ostream &os, node<T> * &head )
{
    for ( const node<T> *current = head; current != nullptr; current = current->next )
    {
        os << current->data << " -> ";
    }

    return os << "null";
}

int main() 
{
    node<int> *head = nullptr;
    int a[] = { 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 };
    const size_t N = sizeof( a ) / sizeof( *a );

    assign( head, std::begin( a ), std::end( a ) );

    std::cout << head << '\n';

    for ( size_t i = 0, n = 1; i < N; i += n )
    {
        head = append_n( head, n );

        std::cout << head << '\n';
    }

    std::cout << '\n';

    for ( size_t i = 0, n = 2; i < N; i += n )
    {
        head = append_n( head, n );

        std::cout << head << '\n';
    }

    std::cout << '\n';

    for ( size_t i = 0, n = 5; i < N; i += n )
    {
        head = append_n( head, n );

        std::cout << head << '\n';
    }

    std::cout << '\n';

    return 0;
}

程序輸出與上圖相同。

0 -> 1 -> 2 -> 3 -> 4 -> 5 -> 6 -> 7 -> 8 -> 9 -> null
9 -> 0 -> 1 -> 2 -> 3 -> 4 -> 5 -> 6 -> 7 -> 8 -> null
8 -> 9 -> 0 -> 1 -> 2 -> 3 -> 4 -> 5 -> 6 -> 7 -> null
7 -> 8 -> 9 -> 0 -> 1 -> 2 -> 3 -> 4 -> 5 -> 6 -> null
6 -> 7 -> 8 -> 9 -> 0 -> 1 -> 2 -> 3 -> 4 -> 5 -> null
5 -> 6 -> 7 -> 8 -> 9 -> 0 -> 1 -> 2 -> 3 -> 4 -> null
4 -> 5 -> 6 -> 7 -> 8 -> 9 -> 0 -> 1 -> 2 -> 3 -> null
3 -> 4 -> 5 -> 6 -> 7 -> 8 -> 9 -> 0 -> 1 -> 2 -> null
2 -> 3 -> 4 -> 5 -> 6 -> 7 -> 8 -> 9 -> 0 -> 1 -> null
1 -> 2 -> 3 -> 4 -> 5 -> 6 -> 7 -> 8 -> 9 -> 0 -> null
0 -> 1 -> 2 -> 3 -> 4 -> 5 -> 6 -> 7 -> 8 -> 9 -> null

8 -> 9 -> 0 -> 1 -> 2 -> 3 -> 4 -> 5 -> 6 -> 7 -> null
6 -> 7 -> 8 -> 9 -> 0 -> 1 -> 2 -> 3 -> 4 -> 5 -> null
4 -> 5 -> 6 -> 7 -> 8 -> 9 -> 0 -> 1 -> 2 -> 3 -> null
2 -> 3 -> 4 -> 5 -> 6 -> 7 -> 8 -> 9 -> 0 -> 1 -> null
0 -> 1 -> 2 -> 3 -> 4 -> 5 -> 6 -> 7 -> 8 -> 9 -> null

5 -> 6 -> 7 -> 8 -> 9 -> 0 -> 1 -> 2 -> 3 -> 4 -> null
0 -> 1 -> 2 -> 3 -> 4 -> 5 -> 6 -> 7 -> 8 -> 9 -> null