链表的递归递增和递归降序函数

时间:2014-03-06 04:44:38

标签: c++ sorting recursion linked-list

这是一个家庭作业,我已经编写了大部分代码,我唯一想知道的是我必须有一个递归函数来按升序对链表进行排序,以及一个递归函数按降序对链表进行排序。我很丢失。

这是我的整个代码。

using namespace std;

struct ListNode;
typedef ListNode* ListPtr;

struct ListNode
{
    int number;
    ListPtr next;

    ListNode(int value, ListPtr ptr = NULL)
    {
        number = value;
        next = ptr;
    }
};

char Menu();
void Add(ListPtr &, int);
void Delete(ListPtr &, int);
void Ascend(ListPtr &);
void Descend(ListPtr &);
void Print(ListPtr &);
void DeleteList(ListPtr &);

int main()
{
    ListPtr head = NULL;
    char answer;
    int input;

    answer = Menu();
    while(answer != 'Q')
    {
        if(answer == 'A')
        {
            cout << "Please enter in an integer: ";
            cin >> input;
            Add(head, input);
        }
        else if(answer == 'D')
        {
            cin >> input;
            Delete(head, input);
        }
        else if(answer == 'P')
        {
            Ascend(head);
        }
        else if(answer == 'O')
        {
            Descend(head);
        }
        else if(answer == 'N')
        {
            Print(head);
        }
        else
        {
            cout << "Incorrect input, please try again.\n";
        }

        answer = Menu();
    }

    DeleteList(head);
    return 0;
}

char Menu()
{
    char uinput;

    cout << "Please enter in one of the following:\n";
    cout << "A: Add an item to the end of the list.\n";
    cout << "D: Delete an item from the list.\n";
    cout << "P: Print the list in ascending order.\n";
    cout << "O: Print the list in descending order.\n";
    cout << "N: Display the number of items in the list.\n";
    cout << "Q: Quit.\n";

    return toupper(uinput);
}

void Add(ListPtr &start, int item)
{
    if(start->number > item || start == NULL)
        start = new ListNode(item, start);
    else
        Add(start->next, item);
}

void Delete(ListPtr &start, int item)
{
    if(start != NULL)
    {
        if(start->number == item)
            ListPtr cur = start;
        start = start->next;
        delete cur;
    }
    else
    {
        Delete(start->next, item);
    }
}

void Ascend(ListPtr &start)
{

} 

void Descend(ListPtr &start)
{

}

void Print(ListPtr &start)
{
    ListPtr cur = start;
    int count = 0;

    if(cur == NULL)
    {
        cout << "The list is empty.\n";
    }
    else
    {
        if(cur != NULL)
        {
            if(count % 10 == 0)
                cout << endl;
            cout << setw(5) << cur->number;
            cur = cur->next;
            count++;
        }
    }
    cout << endl;
}


void DeleteList(ListPtr &start)
{
    if(start != NULL)
    {
        DeleteList(start->next);
        cout << "Deleting item " << start->number << endl;
        delete start;
    }
}

1 个答案:

答案 0 :(得分:0)

对于递归部分,一种方法是递归地将列表分成左右两个列表,直到列表大小减少到1(或为零),在这种情况下,递归函数只返回列表,否则它&# 39; s后跟递归函数中的代码,该函数合并返回的左侧和右侧列表,并返回合并列表。

您是否学会了如何将链表拆分为两个列表?通常这是使用两个指针完成的。我不确定你所教的是什么以及你应该弄清楚什么。这是什么级别的编程类?