精华内容
下载资源
问答
  • 葛立恒数太大了需要用高德纳箭头表示。 (一)、高德纳箭头的定义为: (二)、葛立恒数的定义为: ...直观解释葛立恒数: ...宇宙的大小为920亿光年=8*10^26米 ...(三)、TREE(3): 直观介绍:http://www.360...

    葛立恒数太大了需要用高德纳箭头表示。

    (一)、高德纳箭头的定义为:

    (二)、葛立恒数的定义为:

    直观解释葛立恒数:

    宇宙的大小为920亿光年=8*10^26米

    宇宙最短长度为普朗克长度=1.6*10-34米

    宇宙中的普朗克长度约等于10^183  <<  3↑↑↑↑3  << G

     

    (三)、TREE(3):

    直观介绍:http://www.360doc.com/content/18/0806/18/11935121_776161545.shtml

    在TREE(3)面前,葛立恒数可以忽略不计

     

    (四)、SCG3:

    暂时没有找到中文的说明

     

    (五)、SSCG3:

    暂时没有找到中文的说明

     

    (六)、Rayo数(拉约数)

    由阿古斯丁·巴列卡诺(Agustín Rayo)2007年所创造并命名的大数。维基百科说明:http://googology.wikia.com/wiki/Rayo%27s_number

     

    (七)、Fish number 7

    2013年被定义。维基百科说明:http://googology.wikia.com/wiki/Fish_number_7

     

    (八)、BigFoot

    2014年10月被定义。维基百科说明:http://googology.wikia.com/wiki/BIG_FOOT

     

    (九)、Little Bigeddon

    2017年1月5日被定义。维基百科说明:http://googology.wikia.com/wiki/Little_Bigeddon

     

    (十)、Sasquatch (Big Bigeddon)

    2017年3月27号被定义。维基百科说明:http://googology.wikia.com/wiki/Sasquatch

    创造者的博客说明:http://googology.wikia.com/wiki/User_blog:Emlightened/Sasquatch_(Big_Bigeddon)

     

     

    上述大数均是有界的,即都存在上限,可以写为一个确切的数(如果能够写下的话)。

    接下来介绍几个无界的,也就是无穷的。

     

    (十一)、阿列夫零

    通俗解释:所有整数构成的集合的基数

     

    (十二)、阿列夫一

    通俗解释:所有实数构成的集合的基数

     

    (十三)阿列夫三

     

    (十四)等等

     

    (十五)∞

     

     

    展开全文
  • sqlite3BtreeClearTable:Removes all data from a B-tree, but keeps the B-tree intact. 2.5.6.3 游标函数(CursorFunctions) sqlite3BtreeCursor:Creates a new cursor pointing to a particular B-tree. ...

    2.5.3 页面重用及回收(PageReuse and Vacuum )

    SQLite利用一个freelist进行页面回收。当一个页面的所有记录都被删除时,就被插入到该列表。当运行VACUUM命令时,会清除freelist,所以数据库会缩小,本质上它是在新的文件重新建立数据库,而所有使用的页在都被拷贝过去,而freelist却不会,结果就是一个新的,变小的数据库。当数据库的autovacuum开启时,SQLite不会使用freelist,而且在每一次commit时自动压缩数据库。

    2.5.4 B-Tree记录

    B-tree中页面由B-tree记录组成,也叫做payloads。每一个B-tree记录,或者payload有两个域:关键字域(keyfield)和数据域(data field)。Key field就是rowid的值,或者数据库中表的关键字的值。从B-tree的角度,data field可以是任何无结构的数据。数据库的记录就保存在这些data fields中。B-tree的任务就是排序和遍历,它最需要就是关键字。Payloads的大小是不定的,这与内部的关键字和数据域有关,当一个 payload太大不能存在一个页面内进便保存到多个页面。

    B-tree按关键字排序,所有的关键字必须唯一。表采用B+tree,内部页面不包含数据。图2.10展示了一个利用B+tree表示表的例子。

    B+tree中根页面(rootpage)和内部页面(internal pages)都是用来导航的,这些页面的数据域都是指向下级页面的指针,仅仅包含关键字。所有的数据库记录都存储在叶子页面(leaf pages)内。在叶节点一级,记录和页面都是按照关键字的顺序的,所以B-tree可以水平方向遍历,时间复杂度为O(1)。

    图2.10 B+tree结构(tables)

    2.5.5 记录和域(Recordsand Fields)

    位于叶节点页面的数据域的记录由VDBE管理,数据库记录以二进制的形式存储,但有一定的数据格式。记录格式包括一个逻辑头(logical header)和一个数据区(data segment),header segment包括header的大小和一个数据类型数组,数据类型用来表示存储在data segment的数据的类型,如图2.11所示。

    图2.11 记录结构

    2.5.6 B-Tree API

    B-Tree模块有它自己的API,它可以独立于C API使用。另一个特点就是它支持事务。由pager处理的事务,锁和日志都是为B-tree服务的。

    2.5.6.1 访问和事务函数

    sqlite3BtreeOpen:Opens a new database file. Returns a B-tree object.

    sqlite3BtreeClose:Closes a database.

    sqlite3BtreeBeginTrans:Starts a new transaction.

    sqlite3BtreeCommit:Commits the current transaction.

    sqlite3BtreeRollback:Rolls back the current transaction.

    sqlite3BtreeBeginStmt:Starts a statement transaction.

    sqlite3BtreeCommitStmt:Commits a statement transaction.

    sqlite3BtreeRollbackStmt:Rolls back a statement transaction.

    2.5.6.2 表函数

    sqlite3BtreeCreateTable:Creates a new, empty B-tree in a database file.

    sqlite3BtreeDropTable:Destroys a B-tree in a database file.

    sqlite3BtreeClearTable:Removes all data from a B-tree, but keeps the B-tree intact.

    2.5.6.3 游标函数(CursorFunctions)

    sqlite3BtreeCursor:Creates a new cursor pointing to a particular B-tree.

    sqlite3BtreeCloseCursor:Closes the B-tree cursor.

    sqlite3BtreeFirst:Moves the cursor to the first element in a B-tree.

    sqlite3BtreeLast:Moves the cursor to the last element in a B-tree.

    sqlite3BtreeNext:Moves the cursor to the next element after the one it is currently

    pointingto.

    sqlite3BtreePrevious:Moves the cursor to the previous element before the one it is

    currentlypointing to.

    sqlite3BtreeMoveto:Moves the cursor to an element that matches the key value passed in as aparameter.

    2.5.6.4 记录函数(RecordFunctions)

    sqlite3BtreeDelete:Deletes the record that the cursor is pointing to.

    sqlite3BtreeInsert:Inserts a new element in the appropriate place of the B-tree.

    sqlite3BtreeKeySize:Returns the number of bytes in the key of the record that the

    cursoris pointing to.

    sqlite3BtreeKey:Returns the key of the record the cursor is currently pointing to.

    sqlite3BtreeDataSize:Returns the number of bytes in the data record that the cursor is

    currentlypointing to.

    sqlite3BtreeData:Returns the data in the record the cursor is currently pointing to.

    2.5.6.5 配置函数(ConfigurationFunctions)

    sqlite3BtreeSetCacheSize:Controls the page cache size as well as the synchronous

    writes(as defined in the synchronous pragma).

    sqlite3BtreeSetSafetyLevel:Changes the way data is synced to disk in order to increase

    ordecrease how well the database resists damage due to OS crashes and powerfailures.

    Level 1is the same as asynchronous (no syncs() occur and there is a high probabilityof

    damage).This is the equivalent to pragma synchronous=OFF. Level 2 is the default. There

    is avery low but non-zero probability of damage. This is the equivalent to pragma

    synchronous=NORMAL.Level 3 reduces the probability of damage to near zero but with a

    writeperformance reduction. This is the equivalent to pragma synchronous=FULL.

    sqlite3BtreeSetPageSize:Sets the database page size.

    sqlite3BtreeGetPageSize:Returns the database page size.

    sqlite3BtreeSetAutoVacuum:Sets the autovacuum property of the database.

    sqlite3BtreeGetAutoVacuum:Returns whether the database uses autovacuum.

    sqlite3BtreeSetBusyHandler:Sets the busy handler

    展开全文
  • ~tree(); node *getRoot(){ return root;} int Depth(node *); void PreOrder(node * ); void InOrder(node *); void PostOrder(node *); void LayerTravel(node *); void PrintLevel(node *...

    1.二叉树的实现

    按前序遍历进行输入,#表示空树

    struct node{
        char data;
        struct node *left, *right;
    };
    
    class tree{
    public:
        tree();
        ~tree();
        node *getRoot(){return root;}
    
        int Depth(node *);
    
        void PreOrder(node * );
        void InOrder(node *);
        void PostOrder(node *);
        void LayerTravel(node *);
        void PrintLevel(node *, int);
    
        void NonrePreOrder(node *);
        void NonreInOrder(node *);
        void NonrePostOrder(node *);
        void NonreLayerTravel(node *);
    
    private:
        node* root;
        node* creat();
    };

    构造函数

    tree::tree(){
        this->root = creat();
    }

    creat()方法

    node* tree::creat(){
        node* root;
        char ch;
        cout<<"Input:"<<endl;
        cin>>ch;
        if(ch == '#')root = NULL;
        else{
            root = new node;
            root->data = ch;
            root->left = creat();
            root ->right = creat();
        }
        return root;
    }

    2.二叉树的遍历

    1.递归前序遍历

    void tree::PreOrder(node* root){
         if(root!=NULL){
            cout<<root->data<<" ";
            PreOrder(root->left);
            PreOrder(root->right);
         }
    }

    2.递归中序遍历

    void tree::InOrder(node* root){
         if(root!=NULL){
            InOrder(root->left);
            cout<<root->data<<" ";
            InOrder(root->right);
         }
    }

    3.递归后序遍历

    void tree::PostOrder(node* root){
         if(root!=NULL){
            PostOrder(root->left);
            PostOrder(root->right);
            cout<<root->data<<" ";
         }
    }

    4.递归层次遍历

    递归求树的高度

    int tree::Depth(node *root){
        if(root==NULL)return 0;
        return 1 + max(Depth(root->left),Depth(root->right));
    }
    void tree::LayerTravel(node* root){
         if(root!=NULL){
            int depth = Depth(root);
            int i;
            for(i=1;i<=depth;i++){
             PrintLevel(root,i);
             cout<<endl;
            }
         }
    }
    void tree::PrintLevel(node* root,int level){
        if(root == NULL || level < 1) return;
        if(1==level){
            cout<<root->data<<" ";
            return;
        }
        PrintLevel(root->left,level-1);
        PrintLevel(root->right,level-1);
    }

    5.非递归前序遍历

    void tree::NonrePreOrder(node* root){
        stack<node*> s;
        node* p = root;
        do{
            while(p){
                cout<<p->data<<" ";
                s.push(p);
                p = p->left;
            }
            if(!s.empty()){
                p = s.top();
                s.pop();
                p = p->right;
            }
        }while(p||!s.empty());
    }

    6.非递归中序遍历

    void tree::NonreInOrder(node* root){
        stack<node*> s;
        node* p = root;
        do{
            while(p){s.push(p); p = p->left;}
            if(!s.empty()){
                p=s.top();
                s.pop();
                cout<<p->data<<" ";
                p = p->right;
            }
        }while(p||!s.empty());
    }

    7.非递归后序遍历

    这里写图片描述
    这里写图片描述

    void tree::NonrePostOrder(node* root){
       if (root == NULL) return;
        stack<node*> s;
        s.push(root);
        node* head = root;
        while (!s.empty()){
            node* t = s.top();
            if (!t->left&&!t->right || t->left == head || t->right == head){
                cout<<t->data<<" ";
                s.pop();
                head = t;
            }
            else{
                if (t->right)s.push(t->right);
                if (t->left)s.push(t->left);
            }
    
        }
    }

    8.非递归层次遍历

    void tree::NonreLayerTravel(node* root){
    
        if (root == NULL)return;
        queue<node*> q;
        q.push(root);
        while (!q.empty()){
            //计算当前队列中含有多少元素
            int size = q.size();
            //从头部依次取出size个元素
            for (int i = 0; i < size; i++){
                root = q.front();
                q.pop();
                cout<<root->data<<" ";
                //将下一层的元素塞入队列
                if (root->left)q.push(root->left);
                if (root->right)q.push(root->right);
            }
        }
    }

    3.完整代码

    头文件a.h

    #ifndef A_H_INCLUDED
    #define A_H_INCLUDED
    #include <iostream>
    #include <stack>
    #include <queue>
    using namespace std;
    /
    struct node{
        char data;
        struct node *left, *right;
    };
    
    class tree{
    public:
        tree();
        ~tree();
        node *getRoot(){return root;}
    
        int Depth(node *);
    
        void PreOrder(node * );
        void InOrder(node *);
        void PostOrder(node *);
        void LayerTravel(node *);
        void PrintLevel(node *, int);
    
        void NonrePreOrder(node *);
        void NonreInOrder(node *);
        void NonrePostOrder(node *);
        void NonreLayerTravel(node *);
    
    private:
        node* root;
        node* creat();
    };
    
    tree::tree(){
        this->root = creat();
    }
    
    
    node* tree::creat(){
        node* root;
        char ch;
        cout<<"Input:"<<endl;
        cin>>ch;
        if(ch == '#')root = NULL;
        else{
            root = new node;
            root->data = ch;
            root->left = creat();
            root ->right = creat();
        }
        return root;
    }
    int tree::Depth(node *root){
        if(root==NULL)return 0;
        return 1 + max(Depth(root->left),Depth(root->right));
    }
    
    void tree::PreOrder(node* root){
         if(root!=NULL){
            cout<<root->data<<" ";
            PreOrder(root->left);
            PreOrder(root->right);
         }
    }
    
    void tree::InOrder(node* root){
         if(root!=NULL){
            InOrder(root->left);
            cout<<root->data<<" ";
            InOrder(root->right);
         }
    }
    
    void tree::PostOrder(node* root){
         if(root!=NULL){
            PostOrder(root->left);
            PostOrder(root->right);
            cout<<root->data<<" ";
         }
    }
    
    void tree::LayerTravel(node* root){
         if(root!=NULL){
            int depth = Depth(root);
            int i;
            for(i=1;i<=depth;i++){
             PrintLevel(root,i);
             cout<<endl;
            }
         }
    }
    
    void tree::PrintLevel(node* root,int level){
        if(root == NULL || level < 1) return;
        if(1==level){
            cout<<root->data<<" ";
            return;
        }
        PrintLevel(root->left,level-1);
        PrintLevel(root->right,level-1);
    
    }
    
    
    void tree::NonrePreOrder(node* root){
        stack<node*> s;
        node* p = root;
        do{
            while(p){
                cout<<p->data<<" ";
                s.push(p);
                p = p->left;
            }
            if(!s.empty()){
                p = s.top();
                s.pop();
                p = p->right;
            }
        }while(p||!s.empty());
    }
    
    void tree::NonreInOrder(node* root){
        stack<node*> s;
        node* p = root;
        do{
            while(p){s.push(p); p = p->left;}
            if(!s.empty()){
                p=s.top();
                s.pop();
                cout<<p->data<<" ";
                p = p->right;
            }
        }while(p||!s.empty());
    }
    
    void tree::NonrePostOrder(node* root){
       if (root == NULL) return;
        stack<node*> s;
        s.push(root);
        node* head = root;
        while (!s.empty()){
            node* t = s.top();
            if (!t->left&&!t->right || t->left == head || t->right == head){
                cout<<t->data<<" ";
                s.pop();
                head = t;
            }
            else{
                if (t->right)s.push(t->right);
                if (t->left)s.push(t->left);
            }
    
        }
    }
    
    void tree::NonreLayerTravel(node* root){
    
        if (root == NULL)return;
        queue<node*> q;
        q.push(root);
        while (!q.empty()){
            //计算当前队列中含有多少元素
            int size = q.size();
            //从头部依次取出size个元素
            for (int i = 0; i < size; i++){
                root = q.front();
                q.pop();
                cout<<root->data<<" ";
                //将下一层的元素塞入队列
                if (root->left)q.push(root->left);
                if (root->right)q.push(root->right);
            }
        }
    }
    /
    #endif // A_H_INCLUDED

    测试代码main.cpp

    #include "a.h"
    #include <iostream>
    using namespace std;
    int main(){
    
        tree *mytree = new tree();
        cout<<"Creat sucessfully!!"<<endl;
        node* myroot = mytree->getRoot();
        cout<<"-------------------------------------"<<endl;
        cout<<"The depth is:"<<mytree->Depth(myroot)<<endl;
        cout<<"-------------------------------------"<<endl;
        cout<<"Recursive Traversal"<<endl;
        cout<<" Preorder Traversal:  ";
        mytree->PreOrder(myroot);
        cout<<endl;
        cout<<"  Inorder Traversal:  ";
        mytree->InOrder(myroot);
        cout<<endl;
        cout<<"Postorder Traversal:  ";
        mytree->PostOrder(myroot);
        cout<<endl;
        cout<<"    Layer Traversal:  ";
        cout<<endl;
        mytree->LayerTravel(myroot);
        cout<<endl;
        cout<<"-------------------------------------"<<endl;
        cout<<"Nonrecursive Traversal"<<endl;
        cout<<" Preorder Traversal:  ";
        mytree->NonrePreOrder(myroot);
        cout<<endl;
        cout<<"  Inorder Traversal:  ";
        mytree->NonreInOrder(myroot);
        cout<<endl;
        cout<<"Postorder Traversal:  ";
        mytree->NonrePostOrder(myroot);
        cout<<endl;
        cout<<"    Layer Traversal:  ";
        mytree->NonreLayerTravel(myroot);
        cout<<endl;
        cout<<"-------------------------------------"<<endl;
        return 0;
    };

    测试结果(要构造如下所示的二叉树)
    这里写图片描述
    按如下进行输入(按前序遍历进行输入,#表示空树)
    这里写图片描述
    结果如图
    这里写图片描述

    展开全文
  • There is a kind of binary tree namedred-black treein the data structure. It has the following 5 properties: (1) Every node is either red or black. (2) The root is black.... (3) All the leaves are NU...

    There is a kind of binary tree named red-black tree in the data structure. It has the following 5 properties:

    • (1) Every node is either red or black.
    • (2) The root is black.
    • (3) All the leaves are NULL nodes and are colored black.
    • (4) Each red node must have 2 black descends (may be NULL).
    • (5) All simple paths from any node x to a descendant leaf have the same number of black nodes.

    We call a non-NULL node an internal node. From property (5) we can define the black-height of a red-black tree as the number of nodes on the simple path from the root (excluding the root itself) to any NULL leaf (including the NULL leaf). And we can derive that a red-black tree with black height H has at least 2​H​​−1 internal nodes.

    Here comes the question: given a positive N, how many distinct red-black trees are there that consist of exactly N internal nodes?

    Input Specification:

    Each input file contains one test case which gives a positive integer N (≤500).

    Output Specification:

    For each case, print in a line the number of distinct red-black tees with N internal nodes. Since the answer may be very large, output the remainder of it divided by 1000000007 please.

    Sample Input:

    5
    

    Sample Output:

    8

    题目大意:红黑树需要满足:

    • (1) 每个节点为黑色或红色
    • (2) 根节点黑色
    • (3) 叶子节点为黑色且为空NULL
    • (4) 红色节点的孩子节点为黑(可以为空)
    • (5) 从根节点到任意叶子节点的简单路径经过的黑色节点数相同

    现在给出一个红黑树非空节点数n,求对应的红黑树有多少种。

    解题思路:动态规划

    常规红黑树均为黑根,现定义一个红根红黑树。不难发现一个红黑树的左右子树均为红黑树,即:

    一个黑根红黑树的左右子树为黑高相同的两个黑根红黑树、两个红根红黑树或者一红一黑

    一个红根红黑树的左右子树为黑高相同的两个黑根红黑树

    定义dpB[n][h]:非空节点为n,高度为h的黑根红黑树的个数

           dpR[n][h]:非空节点为n,高度为h的红根红黑树的个数

    那么有:dpB[i][j] = dpB[i][j] + (dpB[k][j-1]+dpR[k][j])*(dpB[i-1-k][j-1]+dpR[i-1-k][j])   k=0,1...i-1
                  dpR[i][j] = dpR[i][j] + (dpB[k][j-1] * dpB[i-1-k][j-1])   k=0,1...i-1

    最终我们要求的结果为dpB[n][0] + dpB[n][1] +...+ dpB[n][maxn]

    #include <bits/stdc++.h>
    using namespace std;
    const int maxn = 505;
    const int mod = 1000000007;
    int n, low, high;
    long long dpB[maxn][maxn]={}, dpR[maxn][maxn]={}, ans = 0;
    int main(){
        scanf("%d", &n);
        dpB[1][1] = 1, dpB[2][1] = 2;
        dpR[1][1] = 1;
        for(int i = 3; i <= n; ++ i){
            low = log2(i+1)/2, high = log2(i+1)*2;
            for(int j = low; j <= high; ++ j){
                for(int k = 1; k < i-1; ++ k){
                    dpB[i][j] = (dpB[i][j] + (((dpB[k][j-1]+dpR[k][j])*(dpB[i-1-k][j-1]+dpR[i-1-k][j]))%mod))% mod;
                    dpR[i][j] = (dpR[i][j] + (dpB[k][j-1] * dpB[i-1-k][j-1]) % mod) % mod;
                }
            }
        }
        for(int i = 0; i < 30; ++ i)
            ans = (ans + dpB[n][i]) % mod;
        printf("%lld", ans);
    }

     

    展开全文
  • 在PTA测试中完全通过了。...def RL_Rotation(Tree): B = Tree.right C = B.left 在旋转函数中,一开始我把B和C都使用自定义的Node类定义了, #RL_Rotation函数 def RL_Rotation(Tree): B = Node(Tree.right) ...
  • There is a kind of balanced binary search tree named red-black tree in the data structure. It has the following 5 properties: (1) Every node is either red or black. (2) The root is black. ...(3) Every
  • There is a kind of balanced binary search tree named red-black tree in the data structure. It has the following 5 properties: (1) Every node is either red or black. (2) The root is black....(3) Every lea
  • [国家集训队2012]tree(陈立杰)

    千次阅读 2017-08-13 19:49:44
    1764. [国家集训队2012]tree(陈立杰) ★★★ 输入文件:nt2012_tree.in 输出文件:nt2012_tree.out 简单对比 时间限制:3 s 内存限制:1024 MB tree(陈立杰) 时间限制:3.0s 内存限制:1.0GB 【大意】 给你...
  • pat甲级1135 Is It A Red-Black Tree题解 欢迎访问我的pat甲级... 题目描述 题意分析 红黑树是一种平衡二叉排序树,它满足以下要求: (1)每个节点都是红色或黑色。...(3)每个叶子(NULL)都是黑色的。 (...
  • There is a kind of balanced binary search tree named red-black tree in the data structure... It has the following 5 properties:(1) Every node is either red or black.(2) The root is black.(3) Every lea...
  • 1135. Is It A Red-Black Tree (30)-PAT甲级真题

    千次阅读 热门讨论 2017-09-20 09:54:34
    There is a kind of balanced binary search tree named red-black tree in the data structure. It has the following 5 properties...(1) Every node is either red or black.(2) The root is black.(3) Every lea...
  • 二项队列(binomial tree)以最坏时间O(logN)支持以上操作,并且插入操作平均花费常数时间。 二项队列不是一棵树,而是树的集合,称为森林,这里的树有特定的形式,同时也具有堆序性,叫做二项树(binomial tree...
  • There is a kind of balanced binary search tree named red-black tree in the data structure. It has the following 5 properties: (1) Every node is either red or black. (2) The root is black....(3) Eve
  • There is a kind of balanced binary search tree named red-black tree in the data ... It has the following 5 properties:(1) Every node is either red or black.(2) The root is black.(3) Every leaf ...
  • 04-树3. Root of AVL Tree (25) 时间限制 100 ms 内存限制 65536 kB 代码长度限制 8000 B 判题程序 Standard 作者 CHEN, Yue An AVL tree is a self-balancing binary se
  • Device Tree(三):代码分析 作者:linuxer 发布于:2014-6-6 16:03 分类:统一设备模型 一、前言 Device Tree总共有三篇,分别是: 1、为何要引入Device Tree,这个机制是用来解决什么问题的?(请...
  • Scikit-learn——Decision Tree(Classifier)

    千次阅读 2018-01-08 20:36:48
    3.使用代价仅为训练样本数取对数; 4.既能够处理数值型数据集(numerical)也能够处理标签型(categorical)数据集 5.能够处理多输出问题; 6.能够用布尔逻辑来表示一种情况的决策过程,不像神经网络完全黑箱; 7.性
  • 3. Add a GROUP BY and aggregate functions to these basic queries and you have hierarchical reports. For example, the total salaries that each employee controls: 给这些基本查询添加GROUP BY(关键字)和...
  • (3)最临近模块:sklearn.neighbors (4)高斯过程模块:sklearn.gaussian_process (5)朴素贝叶斯模块:sklearn.naive_bayes (6)决策树模块:sklearn.tree (7)集成模块:sklearn.ensemble
  • 树(Tree)——(五)搜索二叉树的节点删除和销毁

    万次阅读 多人点赞 2018-08-13 14:17:30
    若为叶子节点则直接删除,如左图节点1,3 ,8或者右图的1,4,8。(若为单独一个根叶子要单独处理) 若为单独一个根叶子 第二种情况: 若该节点,有一个节点,左或是右。因为只有一个节点,直接令祖父节点....
  • CentOS7下载+VM上安装(手动分区)图文步骤详解(2021) 文章目录CentOS7下载+VM上安装(手动分区)图文步骤详解(2021)CentOS7下载官网下载国内镜像下载VM上...3. 往下拉找到Base Distribution,选择自己的版本,点击Tree
  • * Minimal Steiner Tree * G(V, E), A是V的一个子集, 求至少包含A中所有点的最小子树. * 时间复杂度:O(N^3+N*2^A*(2^A+N)) * INIT: d[][]距离矩阵; id[]置为集合A中点的标号; * CALL: steiner(int n, int a); ...
  • Graph Valid Tree题目解法1:DFS解法2:BFS解法3:并查集(union find)参考链接: 题目 Given n nodes labeled from 0 to n-1 and a list of undirected edges (each edge is a pair of nodes), write a function to...
  • 本文就对Boosting家族中一个重要的算法梯度提升树(Gradient Boosting Decison Tree, 以下简称GBDT)做一个总结。GBDT有很多简称,有GBT(Gradient Boosting Tree), GTB(Gradient Tree Boosting ), GBRT...
  • 翻译给定一个二叉树,检查它是否是它本身的镜像(也就是说,中心对称)。例如,这个二叉树是对称的: 1 / \ 2 2 / \ / \ 3 4 4 3但是下面这个却不是: ...原文Given a binary tree, check whet
  • Given a binary tree, check whether it is a mirror of itself (ie, symmetric around its center). ...For example, this binary tree [1,2,2,3,4,4,3] is symmetric: 1 / \ 2 2 / \ / \ 3 4 4 3

空空如也

空空如也

1 2 3 4 5 ... 20
收藏数 61,587
精华内容 24,634
关键字:

tree(3)