Question

我需要使用向量中可用的元素数量来计算可能组合的数量。例如，如果我有一个[0] = 1＆amp; [1] = 1并且layer = 2（树的长度），那么我应该得到01的答案。如果[0] = 2＆amp; [1] = 2并且layer = 2，则答案是00,01或10,11。问题在于矢量中可能存在不同数量的元素，它们的数量也可能不同。我写了一个递归函数，它创建新节点并将它们引用到向量中，每个节点都有这样的向量，但问题是我不知道如何编写PRINT / SHOW有点函数，以显示结果我需要的方式。请提供一些建议或推荐阅读哪些主题。我非常感谢代码建议。

void addNode(vector<int>currentAmount, int layer, int positionNumber,  Node *&myTree)
{
    if (myTree==NULL)
    {
        myTree = new Node;
        myTree->listOfNodes.push_back(new Node);
        myTree->currentAmount=currentAmount;
    }
    if (myTree!=NULL&&layer!=0)
    {
        myTree = new Node;
        myTree->positionNumber=positionNumber;
        myTree->layer=layer;
        currentAmount[positionNumber]-=1;
        for (int i=0; i<currentAmount.size();i++)
        {
            if (currentAmount[i]!=0) {myTree->listOfNodes.push_back(new Node); myTree->positionNumber.push_back(i);}
        }
        for (int i=0; i<myTree->listOfNodes.size(); i++)
            {addNode(currentAmount, layer-1, myTree->positionNumber[i], myTree->listOfNodes[i]);}
    } 
}

void showTree(Node *&Tree)
{
    for (int i=0; i<Tree->listOfNodes.size(); i++)
    {
        showTree(Tree->listOfNodes[i]);
        cout<<Tree->listOfNodes[i]<<' '<<Tree->positionNumber[i]<<<' '<<Tree->layer<<'\t';
    }
}

Answer 1

我将此代码用于B-TREE。

/*

 * C++ Program to Implement B-Tree

 */

#include <iostream>

using namespace std;



// A BTree node

class BTreeNode

{

    private:

        int *keys;

        int t;

        BTreeNode **C;

        int n;

        bool leaf;

    public:

        BTreeNode(int t1, bool leaf1)

        {

            t = t1;

            leaf = leaf1;

            keys = new int[2*t-1];

            C = new BTreeNode *[2*t];

            n = 0;

        }

        // traverse all nodes in a subtree rooted with this node

        void traverse()

        {

            int i;

            for (i = 0; i < n; i++)

            {

                if (leaf == false)

                    C[i]->traverse();

                cout << " " << keys[i];

            }

            if (leaf == false)

                C[i]->traverse();

        }

        void insertNonFull(int k)

        {

            int i = n-1;

            if (leaf == true)

            {

                while (i >= 0 && keys[i] > k)

                {

                    keys[i+1] = keys[i];

                    i--;

                }

                keys[i+1] = k;

                n = n+1;

            }

            else

            {

                while (i >= 0 && keys[i] > k)

                    i--;

                if (C[i+1]->n == 2*t-1)

                {

                    splitChild(i+1, C[i+1]);

                    if (keys[i+1] < k)

                        i++;

                }

                C[i+1]->insertNonFull(k);

            }

        }

        void splitChild(int i, BTreeNode *y)

        {

            BTreeNode *z = new BTreeNode(y->t, y->leaf);

            z->n = t - 1;

            for (int j = 0; j < t-1; j++)

                z->keys[j] = y->keys[j+t];

            if (y->leaf == false)

            {

                for (int j = 0; j < t; j++)

                    z->C[j] = y->C[j+t];

            }

            y->n = t - 1;

            for (int j = n; j >= i+1; j--)

                C[j+1] = C[j];

            C[i+1] = z;

            for (int j = n-1; j >= i; j--)

                keys[j+1] = keys[j];

            keys[i] = y->keys[t-1];

            n = n + 1;

        }

        BTreeNode *search(int k)

        {

            int i = 0;

            while (i < n && k > keys[i])

                i++;

            if (keys[i] == k)

                return this;

            if (leaf == true)

                return NULL;

            return C[i]->search(k);

        }

        friend class BTree;

};



// Class BTree

class BTree

{

    private:

        BTreeNode *root;

        int t;

    public:

        BTree(int _t)

        {

            root = NULL;

            t = _t;

        }

        void traverse()

        {

            if (root != NULL)

                root->traverse();

        }

        BTreeNode* search(int k)

        {

            return (root == NULL)? NULL : root->search(k);

        }

        void insert(int k)

        {

            if (root == NULL)

            {

                root = new BTreeNode(t, true);

                root->keys[0] = k;

                root->n = 1;

            }

            else

            {

                if (root->n == 2*t-1)

                {

                    BTreeNode *s = new BTreeNode(t, false);

                    s->C[0] = root;

                    s->splitChild(0, root);

                    int i = 0;

                    if (s->keys[0] < k)

                        i++;

                    s->C[i]->insertNonFull(k);

                    root = s;

                }

                else

                    root->insertNonFull(k);

            }

        }

};

// Main

int main()

{

    BTree t(3);

    t.insert(10);

    t.insert(20);

    t.insert(5);

    t.insert(6);

    t.insert(12);

    t.insert(30);

    t.insert(7);

    t.insert(17);

    cout << "Traversal of the constucted tree is ";

    t.traverse();

    cout<<endl;

    int k = 6;

    cout<<k<<" is ";

    (t.search(k) != NULL)? cout << "Present\n" : cout << "Not Present\n";

    k = 15;

    cout<<k<<" is ";

    (t.search(k) != NULL)? cout << "Present\n" : cout << "Not Present\n";

    return 0;

}

用C ++创建n-tree

1 个答案: