((TreeNode<K,V>)e).split(this, newTab, j, oldCap)

As an AI language model, I cannot provide a full context of the code you provided. However, I can provide some information about the method call you mentioned.

Based on the method name and parameters, it seems that the code is related to a hash table implementation. The method call is casting the input object "e" to a TreeNode object, which suggests that it is dealing with a tree-based implementation of the hash table. The split() method is likely used to split a node in the tree when the hash table needs to be resized due to an increase in the number of elements.

Without further context, it is difficult to provide a more detailed explanation of this method call.

相关问题

((TreeNode<K,V>)e).split(this, newTab, j, oldCap)是干嘛

的？

这段代码是Java中HashMap扩容时调用的split方法，作用是将哈希表中某个桶中的链表或红黑树节点分裂成两个链表或红黑树节点，并将它们分别放入新的哈希表中的对应桶中。

具体来说，这段代码中的参数含义如下：

- e：表示当前节点，即需要被分裂的节点；
- newTab：表示新的哈希表，即扩容后的哈希表；
- j：表示当前节点在旧哈希表中的桶位置；
- oldCap：表示旧哈希表的容量。

该方法会在对当前节点进行分裂的同时，顺便将其后继节点也进行分裂，最终返回分裂后的新节点。

代码改进，不使用cout<<endl结构#include <iostream> #include <queue> #include <stack> #include <string> using namespace std; struct Student { string name; int number; int score; }; struct TreeNode { Student data; TreeNode* left; TreeNode* right; TreeNode(Student s) : data(s), left(nullptr), right(nullptr) {} }; void postOrder(TreeNode* root) { if (!root) return; postOrder(root->left); postOrder(root->right); cout << root->data.name << " " << root->data.number << " " << root->data.score << endl; } void preOrder(TreeNode* root) { stack<TreeNode*> s; s.push(root); while (!s.empty()) { TreeNode* node = s.top(); s.pop(); if (node) { cout << node->data.name << " " << node->data.number << " " << node->data.score << endl; s.push(node->right); s.push(node->left); } } } void levelOrder(TreeNode* root) { queue<TreeNode*> q; q.push(root); while (!q.empty()) { TreeNode* node = q.front(); q.pop(); if (node) { cout << node->data.name << " " << node->data.number << " " << node->data.score << endl; q.push(node->left); q.push(node->right); } } } int main() { Student students[] = { {"Lei Zhenzi", 101401, 82}, {"Jiang Ziya", 100032, 90}, {"Ne Zha", 101674, 70}, {"Shen Gongbao", 101982, 87}, {"Jiu Weihu", 107431, 75}, {"Tian Zun", 100001, 98}, {"Tai Yi", 101009, 81}, {"Yang Jian", 101321, 63}, {"Huang Feihu", 101567, 72}, {"Zhou Wang", 108160, 55}, {"Li Jing", 102456, 84}, {"Tu Xingsun", 102245, 65}, }; int n = sizeof(students) / sizeof(Student); TreeNode* root = new TreeNode(students[0]); queue<TreeNode*> q; q.push(root); for (int i = 1; i < n; i += 2) { TreeNode* parent = q.front(); q.pop(); parent->left = new TreeNode(students[i]); q.push(parent->left); if (i + 1 < n) { parent->right = new TreeNode(students[i + 1]); q.push(parent->right); } } cout << "Post-order traversal: " << endl; postOrder(root); cout << "Pre-order traversal: " << endl; preOrder(root); cout << "Level-order traversal: " << endl; levelOrder(root); return 0; }

可以使用'\n'代替endl，这样可以避免刷新缓存区的操作，提高输出效率。修改后的代码如下：

#include <iostream>
#include <queue>
#include <stack>
#include <string>

using namespace std;

struct Student {
    string name;
    int number;
    int score;
};

struct TreeNode {
    Student data;
    TreeNode* left;
    TreeNode* right;
    TreeNode(Student s) : data(s), left(nullptr), right(nullptr) {}
};

void postOrder(TreeNode* root) {
    if (!root) return;
    postOrder(root->left);
    postOrder(root->right);
    cout << root->data.name << " " << root->data.number << " " << root->data.score << '\n';
}

void preOrder(TreeNode* root) {
    stack<TreeNode*> s;
    s.push(root);
    while (!s.empty()) {
        TreeNode* node = s.top();
        s.pop();
        if (node) {
            cout << node->data.name << " " << node->data.number << " " << node->data.score << '\n';
            s.push(node->right);
            s.push(node->left);
        }
    }
}

void levelOrder(TreeNode* root) {
    queue<TreeNode*> q;
    q.push(root);
    while (!q.empty()) {
        TreeNode* node = q.front();
        q.pop();
        if (node) {
            cout << node->data.name << " " << node->data.number << " " << node->data.score << '\n';
            q.push(node->left);
            q.push(node->right);
        }
    }
}

int main() {
    Student students[] = {
        {"Lei Zhenzi", 101401, 82},
        {"Jiang Ziya", 100032, 90},
        {"Ne Zha", 101674, 70},
        {"Shen Gongbao", 101982, 87},
        {"Jiu Weihu", 107431, 75},
        {"Tian Zun", 100001, 98},
        {"Tai Yi", 101009, 81},
        {"Yang Jian", 101321, 63},
        {"Huang Feihu", 101567, 72},
        {"Zhou Wang", 108160, 55},
        {"Li Jing", 102456, 84},
        {"Tu Xingsun", 102245, 65},
    };
    int n = sizeof(students) / sizeof(Student);

    TreeNode* root = new TreeNode(students[0]);
    queue<TreeNode*> q;
    q.push(root);
    for (int i = 1; i < n; i += 2) {
        TreeNode* parent = q.front();
        q.pop();
        parent->left = new TreeNode(students[i]);
        q.push(parent->left);
        if (i + 1 < n) {
            parent->right = new TreeNode(students[i + 1]);
            q.push(parent->right);
        }
    }

    cout << "Post-order traversal: \n";
    postOrder(root);

    cout << "Pre-order traversal: \n";
    preOrder(root);

    cout << "Level-order traversal: \n";
    levelOrder(root);

    return 0;
}