实现算法的Java需要帮助

Question

这算法是如此先进我的基本编程技能，我只是不明白我怎么能实现它。我在一个新的问题张贴这一点，因为我不能继续困扰着谁单独给我的算法，这个在前面的问题的注释部分的家伙。

MaxSet(node) = 1 if "node" is a leaf
MaxSet(node) = Max(1 + Sum{ i=0..3: MaxSet(node.Grandchildren[i]) },  
                       Sum{ i=0..1: MaxSet(node.Children[i])      })

由于太迈赫达德获取的算法。

在这里，我的问题是落实两国总和线的一部分，我该怎么办呢？我需要标记每个节点，该算法选择。这只是在节点级设置为true的“标记”变量。我不明白是它作出决定太选择一个节点？

EDIT包括到目前为止我的代码：

public int maxSet(Posisjon<E> bt){
        if (isExternal(bt)){
            return 1; 
        }
        return Math.max(1 + helper1(bt), helper2(bt));
    }

private int helper1(Posisjon<E> node){
    int tmp = 0; 
    if (hasLeft(node)){
        if(hasLeft((Position<E>)node.leftChild())){
            tmp += maxSet(node.leftChild().leftChild());
        }
        if(hasRight((Position<E>)node.leftChild())){
            tmp += maxSet(node.leftChild().rightChild());
        }
    }
    if(hasRight(node)){
        if(hasLeft((Position<E>)node.rightChild())){
            tmp += maxSet(node.leftChild().leftChild());
        }
        if(hasRight((Position<E>)node.rightChild())){
            tmp += maxSet(node.leftChild().rightChild());
        }
    }
    return tmp; 
}
private int helper2(Posisjon<E> node){
    int tmp = 0; 
    if(hasLeft(node)){
        tmp +=maxSet(node.leftChild());
    }
    if(hasRight(node)){
        tmp +=maxSet(node.rightChild());
    }
    return tmp; 
}

这似乎是工作，现在还剩下什么。是真正标记节点作为选择的？是你，我做到这一点？

---

已更新，代码：

public ArrayList<Posisjon<E>> getSelectionSet(Posisjon<E> bt, ArrayList<Posisjon<E>> s){
        if(bt.marked){
            s.add(bt);
        }
        if(hasLeft(bt)){
            if(hasLeft(bt.leftChild())){
                getSelectionSet(bt.leftChild().leftChild(),s);
            }
            if(hasRight(bt.leftChild())){
                getSelectionSet(bt.leftChild().rightChild(),s);
            }
        }
        if(hasRight(bt)){
            if(hasLeft(bt.rightChild())){
                getSelectionSet(bt.rightChild().leftChild(),s);
            }
            if(hasRight(bt.rightChild())){
                getSelectionSet(bt.rightChild().rightChild(),s);
            }
        }
        return s; 
    }

public int maxSet(Posisjon<E> bt){
        if (bt.visited){
            return bt.computedMax; 
        }
        bt.visited = true; 
        int maxIfCurrentNodeIsSelected = 1 + helper1(bt);
        int maxIfCurrentNodeIsNotSelected = helper2(bt);
        if (maxIfCurrentNodeIsSelected > maxIfCurrentNodeIsNotSelected){
            bt.marked = true; 
            bt.computedMax = maxIfCurrentNodeIsSelected; 
        }else{
            bt.marked = false; 
            bt.computedMax = maxIfCurrentNodeIsNotSelected; 
        }
        return maxSet(bt);
    }

提交后，我会后整个代码此！

Answer 1

您目前有没有memoize的每一次函数的返回值。每次你打电话maxSet时候，你应该检查是否已经计算出的结果还是不行。如果有，就返回。如果你还没有计算它，它存储的地方。否则，你的算法将是低效的。 （这种方法被称为“动态编程”。了解它。）

// pseudocode:
public int maxSet(Posisjon<E> bt){
    if (visited[bt])
        return computedMax[bt];

    visited[bt] = true;        

    // You don't need to manually check for being a leaf
    // For leaves 'maxIfCurrentNodeIsSelected' is always larger.
    int maxIfCurrentNodeIsSelected = 1 + helper1(bt);
    int maxIfCurrentNodeIsNotSelected = helper2(bt);

    if (maxIfCurrentNodeIsSelected > maxIfCurrentNodeIsNotSelected) {
         shouldSelect[bt] = true;
         computedMax[bt] = maxIfCurrentNodeIsSelected;
    } else {
         shouldSelect[bt] = false;
         computedMax[bt] = maxIfCurrentNodeIsNotSelected;
    }
}

public Set getSelectionSet(Posisjon<E> bt, Set s) {
    if (shouldSelect[bt]) {
        s.Add(bt);

        // You should check for nulls, of course
        getSelectionSet(bt.leftChild.leftChild, s);
        getSelectionSet(bt.leftChild.rightChild, s);
        getSelectionSet(bt.rightChild.leftChild, s);
        getSelectionSet(bt.rightChild.rightChild, s);
    } else {
        getSelectionSet(bt.leftChild, s);
        getSelectionSet(bt.rightChild, s);
    }
    return s;
}

在称为呼叫getSelectionSet与根节点和一个空Set作为参数后maxSet。