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eatmelon说:
来一段perl语言
#!/usr/bin/perl
use strict;
use warnings;Define a hash to contain the data
my %data;
Read data from a file
open(my $fh, ") {
chomp $line;
my ($key, $value) = split /=/, $line;
$data{$key} = $value;
}Print the data
foreach my $key (keys %data) {
print "$key=$data{$key}\n";
} -
ChatGPT 说:
eatmelon说:
来一段perl语言
#!/usr/bin/perl
use strict;
use warnings;Define a hash to contain the data
my %data;
Read data from a file
open(my $fh, ") {
chomp $line;
my ($key, $value) = split /=/, $line;
$data{$key} = $value;
}Print the data
foreach my $key (keys %data) {
print "$key=$data{$key}\n";
}@ChatGPT 解释一下这段代码
-
eatmelon说:
ChatGPT 说:
eatmelon说:
来一段perl语言
#!/usr/bin/perl
use strict;
use warnings;Define a hash to contain the data
my %data;
Read data from a file
open(my $fh, ") {
chomp $line;
my ($key, $value) = split /=/, $line;
$data{$key} = $value;
}Print the data
foreach my $key (keys %data) {
print "$key=$data{$key}\n";
}解释一下这段代码
:
这段代码使用 Perl 语言从一个文件中读取数据到一个 hash 中,然后打印出 hash 中的数据。首先使用
open
函数打开文件,然后使用chomp
函数去掉文件中每一行末尾的换行符,再使用split
函数将每一行按照等号=
分割成 key 和 value,最后将 key 和 value 赋值到 hash 中,最后使用foreach
循环打印出 hash 中的数据。 -
eatmelon说:
用C加加实现一个函数Clear,这个函数参数是一个指针,指向一个类型为node的二叉树节点(node类型有三个值,分别代表着这指向左子树,右子树的指针和节点本身的值,当没有左子树和右子树时,指针指向null),这个函数可以将参数及其子树的值全部设为零,该函数不得出现异常
,即使参数指向null也要正常运行。
void Clear(node *root) { if(root == NULL) return; Clear(root->left); Clear(root->right); root->val = 0; root->left = NULL; root->right = NULL; }
-
ChatGPT 说:
eatmelon说:
用C加加实现一个函数Clear,这个函数参数是一个指针,指向一个类型为node的二叉树节点(node类型有三个值,分别代表着这指向左子树,右子树的指针和节点本身的值,当没有左子树和右子树时,指针指向null),这个函数可以将参数及其子树的值全部设为零,该函数不得出现异常
,即使参数指向null也要正常运行。
void Clear(node *root) { if(root == NULL) return; Clear(root->left); Clear(root->right); root->val = 0; root->left = NULL; root->right = NULL; }
@ChatGPT
请再给出node的定义 -
eatmelon说:
ChatGPT 说:
eatmelon说:
用C加加实现一个函数Clear,这个函数参数是一个指针,指向一个类型为node的二叉树节点(node类型有三个值,分别代表着这指向左子树,右子树的指针和节点本身的值,当没有左子树和右子树时,指针指向null),这个函数可以将参数及其子树的值全部设为零,该函数不得出现异常
,即使参数指向null也要正常运行。
void Clear(node *root) { if(root == NULL) return; Clear(root->left); Clear(root->right); root->val = 0; root->left = NULL; root->right = NULL; }
请再给出node的定义
:
struct node { int val; node *left; node *right; }; ``
-
ChatGPT 说:
eatmelon说:
ChatGPT 说:
eatmelon说:
用C加加实现一个函数Clear,这个函数参数是一个指针,指向一个类型为node的二叉树节点(node类型有三个值,分别代表着这指向左子树,右子树的指针和节点本身的值,当没有左子树和右子树时,指针指向null),这个函数可以将参数及其子树的值全部设为零,该函数不得出现异常
,即使参数指向null也要正常运行。
void Clear(node *root) { if(root == NULL) return; Clear(root->left); Clear(root->right); root->val = 0; root->left = NULL; root->right = NULL; }
请再给出node的定义
:
struct node { int val; node *left; node *right; }; ``
@ChatGPT 请再给出插入左子树,右子树的函数
-
eatmelon说:
ChatGPT 说:
eatmelon说:
ChatGPT 说:
eatmelon说:
用C加加实现一个函数Clear,这个函数参数是一个指针,指向一个类型为node的二叉树节点(node类型有三个值,分别代表着这指向左子树,右子树的指针和节点本身的值,当没有左子树和右子树时,指针指向null),这个函数可以将参数及其子树的值全部设为零,该函数不得出现异常
,即使参数指向null也要正常运行。
void Clear(node *root) { if(root == NULL) return; Clear(root->left); Clear(root->right); root->val = 0; root->left = NULL; root->right = NULL; }
请再给出node的定义
:
struct node { int val; node *left; node *right; }; ``
请再给出插入左子树,右子树的函数
void insertLeft(node *root, int item) { node *newNode = new node; newNode->val = item; newNode->left = NULL; newNode->right = NULL; root->left = newNode; } void insertRight(node *root, int item) { node *newNode = new node; newNode->val = item; newNode->left = NULL; newNode->right = NULL; root->right = newNode; } ``
-
ChatGPT 说:
eatmelon说:
ChatGPT 说:
eatmelon说:
ChatGPT 说:
eatmelon说:
用C加加实现一个函数Clear,这个函数参数是一个指针,指向一个类型为node的二叉树节点(node类型有三个值,分别代表着这指向左子树,右子树的指针和节点本身的值,当没有左子树和右子树时,指针指向null),这个函数可以将参数及其子树的值全部设为零,该函数不得出现异常
,即使参数指向null也要正常运行。
void Clear(node *root) { if(root == NULL) return; Clear(root->left); Clear(root->right); root->val = 0; root->left = NULL; root->right = NULL; }
请再给出node的定义
:
struct node { int val; node *left; node *right; }; ``
请再给出插入左子树,右子树的函数
void insertLeft(node *root, int item) { node *newNode = new node; newNode->val = item; newNode->left = NULL; newNode->right = NULL; root->left = newNode; } void insertRight(node *root, int item) { node *newNode = new node; newNode->val = item; newNode->left = NULL; newNode->right = NULL; root->right = newNode; } ``
@ChatGPT 请再给出搜索函数
-
eatmelon说:
ChatGPT 说:
eatmelon说:
ChatGPT 说:
eatmelon说:
ChatGPT 说:
eatmelon说:
用C加加实现一个函数Clear,这个函数参数是一个指针,指向一个类型为node的二叉树节点(node类型有三个值,分别代表着这指向左子树,右子树的指针和节点本身的值,当没有左子树和右子树时,指针指向null),这个函数可以将参数及其子树的值全部设为零,该函数不得出现异常
,即使参数指向null也要正常运行。
void Clear(node *root) { if(root == NULL) return; Clear(root->left); Clear(root->right); root->val = 0; root->left = NULL; root->right = NULL; }
请再给出node的定义
:
struct node { int val; node *left; node *right; }; ``
请再给出插入左子树,右子树的函数
void insertLeft(node *root, int item) { node *newNode = new node; newNode->val = item; newNode->left = NULL; newNode->right = NULL; root->left = newNode; } void insertRight(node *root, int item) { node *newNode = new node; newNode->val = item; newNode->left = NULL; newNode->right = NULL; root->right = newNode; } ``
请再给出搜索函数
-
ChatGPT 说:
eatmelon说:
ChatGPT 说:
eatmelon说:
ChatGPT 说:
eatmelon说:
用C加加实现一个函数Clear,这个函数参数是一个指针,指向一个类型为node的二叉树节点(node类型有三个值,分别代表着这指向左子树,右子树的指针和节点本身的值,当没有左子树和右子树时,指针指向null),这个函数可以将参数及其子树的值全部设为零,该函数不得出现异常
,即使参数指向null也要正常运行。
void Clear(node *root) { if(root == NULL) return; Clear(root->left); Clear(root->right); root->val = 0; root->left = NULL; root->right = NULL; }
请再给出node的定义
:
struct node { int val; node *left; node *right; }; ``
请再给出插入左子树,右子树的函数
void insertLeft(node *root, int item) { node *newNode = new node; newNode->val = item; newNode->left = NULL; newNode->right = NULL; root->left = newNode; } void insertRight(node *root, int item) { node *newNode = new node; newNode->val = item; newNode->left = NULL; newNode->right = NULL; root->right = newNode; } ``
@ChatGPT
请再给出能搜索二叉树中含有特定值的结点的函数 -
eatmelon说:
ChatGPT 说:
eatmelon说:
ChatGPT 说:
eatmelon说:
ChatGPT 说:
eatmelon说:
用C加加实现一个函数Clear,这个函数参数是一个指针,指向一个类型为node的二叉树节点(node类型有三个值,分别代表着这指向左子树,右子树的指针和节点本身的值,当没有左子树和右子树时,指针指向null),这个函数可以将参数及其子树的值全部设为零,该函数不得出现异常
,即使参数指向null也要正常运行。
void Clear(node *root) { if(root == NULL) return; Clear(root->left); Clear(root->right); root->val = 0; root->left = NULL; root->right = NULL; }
请再给出node的定义
:
struct node { int val; node *left; node *right; }; ``
请再给出插入左子树,右子树的函数
void insertLeft(node *root, int item) { node *newNode = new node; newNode->val = item; newNode->left = NULL; newNode->right = NULL; root->left = newNode; } void insertRight(node *root, int item) { node *newNode = new node; newNode->val = item; newNode->left = NULL; newNode->right = NULL; root->right = newNode; } ``
请再给出能搜索二叉树中含有特定值的结点的函数
-
eatmelon说:
请基于红黑树技术写出一段搜索树,语言为C++,不得使用stl标准库
#include #include //红黑树节点结构定义 struct Node { int data; int color; Node* left; Node* right; Node* parent; }; //红黑树类定义 class RBTree { private: Node* root; public: RBTree(); ~RBTree(); void insert(Node* node); void deleteNode(Node* node); Node* search(int data); void leftRotate(Node* node); void rightRotate(Node* node); void insertFixUp(Node* node); void deleteFixUp(Node* node); }; //构造函数 RBTree::RBTree() { root = NULL; } //析构函数 RBTree::~RBTree() { delete root; root = NULL; } //搜索函数 Node* RBTree::search(int data) { Node* current = root; while (current != NULL && data != current->data) { if (data data) current = current->left; else current = current->right; } return current; } //插入函数 void RBTree::insert(Node* node) { Node* current = root; Node* parent = NULL; while (current != NULL) { parent = current; if (node->data data) current = current->left; else current = current->right; } node->parent = parent; if (parent == NULL) root = node; else if (node->data data) parent->left = node; else parent->right = node; node->left = NULL; node->right = NULL; node->color = 0; insertFixUp(node); } //左旋函数 void RBTree::leftRotate(Node* node) { Node* right = node->right; node->right = right->left; if (right->left != NULL) right->left->parent = node; right->parent = node->parent; if (node->parent == NULL) root = right; else if (node == node->parent->left) node->parent->left = right; else node->parent->right = right; right->left = node; node->parent = right; } //右旋函数 void RBTree::rightRotate(Node* node) { Node* left = node->left; node->left = left->right; if (left->right != NULL) left->right->parent = node; left->parent = node->parent; if (node->parent == NULL) root = left; else if (node == node->parent->left) node->parent->left = left; else node->parent->right = left; left->right = node; node->parent = left; } //插入修复函数 void RBTree::insertFixUp(Node* node) { Node* parent = NULL; Node* grandparent = NULL; while ((node->parent != NULL) && (node->parent->color == 0)) { parent = node->parent; grandparent = parent->parent; if (parent == grandparent->left) { Node* uncle = grandparent->right; if (uncle != NULL && uncle->color == 0) { grandparent->color = 0; parent->color = 1; uncle->color = 1; node = grandparent; } else { if (node == parent->right) { node = parent; leftRotate(node); } parent = node->parent; grandparent = parent->parent; grandparent->color = 0; parent->color = 1; rightRotate(grandparent); } } else { Node* uncle = grandparent->left; if (uncle != NULL && uncle->color == 0) { grandparent->color = 0; parent->color = 1; uncle->color = 1; node = grandparent; } else { if (node == parent->left) { node = parent; rightRotate(node); } parent = node->parent; grandparent = parent->parent; grandparent->color = 0; parent->color = 1; leftRotate(grandparent); } } } root->color = 1; } //删除函数 void RBTree::deleteNode(Node* node) { //定义临时节点 Node* temp = node; Node* child; int temp_color = temp->color; if (node->left == NULL) { child = node->right; transplant(node, node->right); } else if (node->right == NULL) { child = node->left; transplant(node, node->left); } else { temp = minimum(node->right); temp_color = temp->color; child = temp->right; if (temp->parent == node) { child->parent = temp; } else { transplant(temp, temp->right); temp->right = node->right; temp->right->parent = temp; } transplant(node, temp); temp->left = node->left; temp->left->parent = temp; temp->color = node->color; } if (temp_color == 1) { deleteFixUp(child); } } //删除修复函数 void RBTree::deleteFixUp(Node* node) { Node *parent, *sibling; while ((node != root) && (node->color == 1)) { parent = node->parent; if (node == parent->left) { sibling = parent->right; if (sibling->color == 0) { sibling->color = 1; parent->color = 0; leftRotate(parent); sibling = parent->right; } if ((sibling->left->color == 1) && (sibling->right->color == 1)) { sibling->color = 0; node = parent; } else { if (sibling->right->color == 1) { sibling->left->color = 1; sibling->color = 0; rightRotate(sibling); sibling = parent->right; } sibling->color = parent->color; parent->color = 1; sibling->right->color = 1; leftRotate(parent); node = root; } } else { sibling = parent->left; if (sibling->color == 0) { sibling->color = 1; parent->color = 0; rightRotate(parent); sibling = parent->left; } if ((sibling->left->color == 1) && (sibling->right->color == 1)) { sibling->color = 0; node = parent; } else { if (sibling->left->color == 1) { sibling->right->color = 1; sibling->color = 0; leftRotate(sibling); sibling = parent->left; } sibling->color = parent->color; parent->color = 1; sibling->left->color = 1; rightRotate(parent); node = root; } } } node->color = 1; }