Question

我正在尝试为绳索数据结构编写拆分绳索函数，以便可以在索引I函数处插入。我找到了从索引开始到结束检索子字符串的Java实现，并且我试图从中使用这些概念来创建C ++函数来分割绳索

我已经引用了Wikipedia文章和原始论文（在下面链接），但是它们非常模糊。 https://en.wikipedia.org/wiki/Rope_(data_structure) http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.14.9450&rep=rep1&type=pdf

这是我引用的代码的链接： https://www.sanfoundry.com/java-program-implement-rope/

string substring(int start, int end){
            string str = "";
            bool found = false;
            Node *temp = root;
            if (end > temp->weight) {
                found = true;
                end -= temp->weight;
                if (start > temp->weight) {
                    start -= temp->weight;
                    str = temp->right->data[substring(start, end)];
                    return str;
                } else
                    str = temp->right->data[substring(0, end)];            
            }        
            if (!found){
                while (end <= temp->weight)
                    temp = temp->left;
                end -= temp->weight;
                if (start >= temp->weight){
                    start -= temp->weight;
                    str = temp->right->data[substring(start, end)] + str;
                    return str;
                }
                str = temp->right->data[substring(0, end)];            
            }    
            temp = temp->left;
            while (start < temp->weight){
                str = temp->right->data + str;
                temp = temp->left;
            }
            start -= temp->weight;
            str = temp->right->data[substring(start, end) + str];    

            return str;        
        }

上面的函数应该从int start到int end返回绳索的子字符串。我收到一个no operator "[]" matches these operands -- operand types are: char * [ std::__1::basic_string<char, std::__1::char_traits<char>, std::__1::allocator<char>> ]错误。

如果可能的话，我想尝试其他类型的实现，但是我找不到关于此功能的更多深入文档

根据要求，这是我编写的其他代码。如果您有关于如何更好地实现拆分成我的代码的建议，将不胜感激

class Rope {   // Rope data structure
private:

    class Node {
    public:
        char *data;
        int weight;
        Node *left, *right;
        Node(char *data) : data(data), weight(sizeof(data)/sizeof(char)), left(nullptr), right(nullptr){}
        Node() : data(nullptr), weight(0), left(nullptr), right(nullptr){}
    };

    Node* root;

public:
    Rope() : root(new Node()) {}

    // Rope(const char str[]) {}
    // Rope(const char str[], int len) {}

    void makeEmpty () {
        root = new Node();
    }

    void concat(char *str) {    // concats string and rope
        Node *temp = new Node(str);
        Node *newRoot = new Node();
        newRoot->left = root;
        newRoot->right = temp;
        newRoot->weight = newRoot->left->weight;
        if (newRoot->left->right != nullptr) {
            newRoot->weight += newRoot->left->right->weight;
        }
        root = newRoot;   
    }

    void concatRope(const Rope& r1, const Rope& r2){ // concats 2 ropes
        Node *newRoot = new Node();
        newRoot->left = r1.root;
        newRoot->right = r2.root;
        newRoot->weight = newRoot->left->weight;
        if (newRoot->left->right != nullptr){
            newRoot->weight += newRoot->left->right->weight;
        }
        root = newRoot;
    }

无法编写C ++ split_rope函数

0 个答案: