Question

考虑这个完全有效的代码：

#include <type_traits>

template <typename T, typename IndexPack> struct Make;

template <typename T, template <T...> class P, T... Indices>
struct Make<T, P<Indices...>> {
    using type = P<(Indices+1)..., (-3*Indices)..., (Indices-1)...>;
};

template <int...> class Pack;

int main() {
    static_assert (std::is_same<Make<int, Pack<1,2,3,4>>::type, 
        Pack<2,3,4,5, -3,-6,-9,-12, 0,1,2,3>>::value, "false");
}

我真正想要的输出是

Pack<2,-3,0, 3,-6,1, 4,-9,2, 5,-12,3>

而不是Pack<2,3,4,5, -3,-6,-9,-12, 0,1,2,3>。我第一次尝试了

using type = P<(Indices+1, -3*Indices, Indices-1)...>;

但编译器简单地将其理解为无用的逗号运算符。得到我想要的东西的理想语法是什么？如果没有这样的语法，那么最简洁的方法是什么，请记住，使用Indices 3次只是一个例子（我们可能想要使用它超过3次）。请不要告诉我，我必须写一个助手来提取单个包，然后“交织”所有元素。这种噩梦般的方法不是最好的解决方案（只有当我们确切地知道要提取多少个包时，这样的解决方案才有效。）

将定义

template <typename T, template <T...> class P, T I>
struct Component {
    using type = P<I+1, -3*I, I-1>;
};

以某种方式提供帮助？对此进行扩展包装？

Answer 1

是的，你可以递归地结束：

public class BinarySearchTree {
    private boolean empty = true;
    private int size = 0;
    private Node root;

    public BinarySearchTree(Node root)
    {
        this.root = root;
    }

    public Node[] search(int value)
    {
        Node parent = null;
        Node currentNode = root;

        Node[] returnList = new Node[2];

        while ( currentNode != null )
        {
            if (currentNode.getValue() == value)
            {
                returnList[0] = parent;
                returnList[1] = currentNode;

                return returnList;
            }
            else if (value < currentNode.getValue())
            {
                parent = currentNode;
                currentNode = currentNode.getLeft();
            }
            else if (value > currentNode.getValue())
            {
                parent = currentNode;
                currentNode = currentNode.getRight();
            }
        }

        return returnList;
    }

    public void add(int value)
    {
        Node comparingNode = root;

        while (true)
        {
            if (comparingNode.getValue() == value)
            {
                System.out.println("Tried to add duplicate value of " + value);
                break;
            }
            if (comparingNode.getLeft() == null && comparingNode.getRight() == null)
            {
                if (value > comparingNode.getValue())
                {
                    comparingNode.setRight(new Node(value));
                }
                if (value < comparingNode.getValue())
                {
                    comparingNode.setLeft(new Node(value));
                }
                break;
            }
            else
            {
                if (value < comparingNode.getValue())
                {
                    if (comparingNode.getLeft() == null)
                    {
                        comparingNode.setLeft(new Node(value));
                        break;
                    }
                    else
                    {
                        comparingNode = comparingNode.getLeft();
                    }
                }
                if (value > comparingNode.getValue())
                {
                    if (comparingNode.getRight() == null)
                    {
                        comparingNode.setRight(new Node(value));
                        break;
                    }
                    else
                    {
                        comparingNode = comparingNode.getRight();
                    }
                }
            }
        }
    }

    public void remove(int value)
    {
        Node[] nodesFound = search( value );
        Node parent = nodesFound[0];
        Node child = nodesFound[1];

        boolean fLeft = ( parent.getLeft() == child );

        // child node has no children.
        if (fLeft)
        {
            parent.setLeft(null);
        }
        else
        {
            parent.setRight(null);          
        }

        if( child.getLeft() != null && child.getRight() == null )
        {
            // child node has only left child.
            if( fLeft )
            {
                parent.setLeft(child.getLeft());
            }
            else
            {
                parent.setRight(child.getLeft());
            }
        }
        else if ( child.getRight() != null && child.getLeft() == null )
        {
            // child node has only right child.
            if( fLeft )
            {
                parent.setLeft(child.getRight());
            }
            else
            {
                parent.setRight(child.getRight());
            }
        }
        else
        {
            // child node has both children.
            if( child.getRight().getLeft() == null )
            {
                child.getRight().setLeft( child.getLeft() );
                parent.setRight( child.getRight() );
            }
            else
            {
                Node[] returnList = findLeftMost2Children(child.getRight());
                Node leftMostParent = returnList[0];
                Node leftMostChild = returnList[1];

                leftMostParent.setLeft(null);

                leftMostChild.setLeft(child.getLeft());
                leftMostChild.setRight(child.getRight());
                parent.setRight(leftMostChild);
            }
        }
    }

    public Node getRoot()
    {
        return root;
    }

    public void outputTreeInOrder( Node root )
    {
        if( root == null )
            return;

        // Output the left tree.
        if( root.getLeft() != null )
            outputTreeInOrder( root.getLeft() );

        // Output the current node.
        System.out.print( root.getValue() + " " );

        // Output the right tree.
        if( root.getRight() != null )
            outputTreeInOrder( root.getRight() );       
    }

    private Node[] findLeftMost2Children( Node root )
    {
        Node parent = null;
        Node current = root;

        while (current.getLeft() != null)
        {
            parent = current;
            current = current.getLeft();
        }

        Node[] returnList = new Node[2];
        returnList[0] = parent;
        returnList[1] = current;

        return returnList;
    }
}

Demo

Answer 2

这是Columbo的解决方案的灵感来源。它使用我最初寻求的包扩展语法，即

using type = typename Merge<T, typename Component<T, P, Indices>::type...>::type;

因此，现在Make可重复使用，首先使用Triple，然后使用Quadruple，因此Indices的任意数量的用法都可以同时展开。此处Component是传递到Make的模板模板模板参数：

#include <type_traits>

template <typename T, typename... Packs> struct Merge;

template <typename T, template <T...> class P1, template <T...> class P2, T... Is, T... Js>
struct Merge<T, P1<Is...>, P2<Js...>> {
    using type = P1<Is..., Js...>;
};

template <typename T, typename Pack1, typename Pack2, typename... Packs>
struct Merge<T, Pack1, Pack2, Packs...> {
    using type = typename Merge<T, Pack1, typename Merge<T, Pack2, Packs...>::type>::type;
};

template <typename T, template <T...> class P, T I>
struct Triple {
    using type = P<I+1, -3*I, I-1>;
};

template <typename T, template <T...> class P, T I>
struct Quadruple {
    using type = P<I+1, -3*I, I-1, I>;
};

template <typename T, typename IndexPack,
    template <typename U, template <U...> class P, U I> class Component> struct Make;

template <typename T, template <T...> class Z, T... Indices,
    template <typename U, template <U...> class P, U I> class Component>
struct Make<T, Z<Indices...>, Component> {
    using type = typename Merge<T, typename Component<T, Z, Indices>::type...>::type;
};

template <int...> class Pack;

int main() {
    static_assert (std::is_same<Make<int, Pack<1,2,3,4>, Triple>::type,
        Pack<2,-3,0, 3,-6,1, 4,-9,2, 5,-12,3>>::value, "false");

    static_assert (std::is_same<Make<int, Pack<1,2,3,4>, Quadruple>::type,
        Pack<2,-3,0,1, 3,-6,1,2, 4,-9,2,3, 5,-12,3,4>>::value, "false");
}

用于几个组件的指数技巧

2 个答案: