给定一个整数数组arr [],任务是从给定数组中删除重复项。
例子:
Input: arr[] = {1, 2, 3, 2, 5, 4, 4}
Output: arr[] = {1, 2, 3, 4, 5}
Input: arr[] = {127, 234, 127, 654, 355, 789, 355, 355, 999, 654}
Output: arr[] = {127, 234, 355, 654, 789, 999}
可以使用二进制搜索树删除数组中的重复项。想法是使用数组元素创建二叉搜索树,条件是第一个元素被视为根(父)元素,并且当元素“小于根”出现时,它成为左子元素,而元素“大于根”是根的正确子代。由于不存在“等于”的条件,因此当我们从数组元素中形成二叉搜索树时,重复项将被自动删除。
For the array, arr[] = {1, 2, 3, 2, 5, 4, 4}
BST will be:
方法:
- 使用数组元素形成BST
- 使用任何“树遍历”方法显示元素。
下面是上述方法的实现。
C++
// C++ Program of above implementation
#include
using namespace std;
// Struct declaration
struct Node {
int data;
struct Node* left;
struct Node* right;
};
// Node creation
struct Node* newNode(int data)
{
struct Node* nn
= new Node;
nn->data = data;
nn->left = NULL;
nn->right = NULL;
return nn;
}
// Function to insert data in BST
struct Node* insert(struct Node* root, int data)
{
if (root == NULL)
return newNode(data);
else {
if (data < root->data)
root->left = insert(root->left, data);
if (data > root->data)
root->right = insert(root->right, data);
return root;
}
}
// InOrder function to display value of array
// in sorted order
void inOrder(struct Node* root)
{
if (root == NULL)
return;
else {
inOrder(root->left);
cout << root->data << " ";
inOrder(root->right);
}
}
// Driver code
int main()
{
int arr[] = { 1, 2, 3, 2, 5, 4, 4 };
// Finding size of array arr[]
int n = sizeof(arr) / sizeof(arr[0]);
struct Node* root = NULL;
for (int i = 0; i < n; i++) {
// Insert element of arr[] in BST
root = insert(root, arr[i]);
}
// Inorder Traversal to print nodes of Tree
inOrder(root);
return 0;
}
// This code is contributed by shivanisingh
C
// C Program of above implementation
#include
#include
// Struct declaration
struct Node {
int data;
struct Node* left;
struct Node* right;
};
// Node creation
struct Node* newNode(int data)
{
struct Node* nn
= (struct Node*)(malloc(sizeof(struct Node)));
nn->data = data;
nn->left = NULL;
nn->right = NULL;
return nn;
}
// Function to insert data in BST
struct Node* insert(struct Node* root, int data)
{
if (root == NULL)
return newNode(data);
else {
if (data < root->data)
root->left = insert(root->left, data);
if (data > root->data)
root->right = insert(root->right, data);
return root;
}
}
// InOrder function to display value of array
// in sorted order
void inOrder(struct Node* root)
{
if (root == NULL)
return;
else {
inOrder(root->left);
printf("%d ", root->data);
inOrder(root->right);
}
}
// Driver code
int main()
{
int arr[] = { 1, 2, 3, 2, 5, 4, 4 };
// Finding size of array arr[]
int n = sizeof(arr) / sizeof(arr[0]);
struct Node* root = NULL;
for (int i = 0; i < n; i++) {
// Insert element of arr[] in BST
root = insert(root, arr[i]);
}
// Inorder Traversal to print nodes of Tree
inOrder(root);
return 0;
}
Java
// Java implementation of the approach
import java.util.Scanner;
// Node declaration
class Node
{
int data;
public Node left;
public Node right;
Node(int data)
{
this.data = data;
left = right = null;
}
}
class GFG
{
// Function to insert data in BST
public static Node insert(Node root, int data)
{
if (root == null)
return new Node(data);
if (data < root.data)
root.left = insert(root.left, data);
if (data > root.data)
root.right = insert(root.right, data);
return root;
}
// InOrder function to display value of array
// in sorted order
public static void inOrder(Node root)
{
if (root == null)
return;
inOrder(root.left);
System.out.print(root.data+" ");
inOrder(root.right);
}
// Driver Code
public static void main(String []args){
int arr[] = { 1, 2, 3, 2, 5, 4, 4 };
// Finding size of array arr[]
int n = arr.length;
Node root = null;
for (int i = 0; i < n; i++)
{
// Insert element of arr[] in BST
root = insert(root,arr[i]);
}
// Inorder Traversal to print nodes of Tree
inOrder(root);
}
}
// This code is contributed by anishma
Python3
# Python3 implementation of the approach
# Binary tree node consists of data, a
# pointer to the left child and a
# pointer to the right child
class newNode :
def __init__(self,data) :
self.data = data;
self.left = None;
self.right = None;
# Function to insert data in BST
def insert(root, data) :
if (root == None) :
return newNode(data);
else :
if (data < root.data) :
root.left = insert(root.left, data);
if (data > root.data) :
root.right = insert(root.right, data);
return root;
# InOrder function to display value of array
# in sorted order
def inOrder(root) :
if (root == None) :
return;
else :
inOrder(root.left);
print(root.data, end = " ");
inOrder(root.right);
# Driver code
if __name__ == "__main__" :
arr = [ 1, 2, 3, 2, 5, 4, 4 ];
# Finding size of array arr[]
n = len(arr);
root = None;
for i in range(n) :
# Insert element of arr[] in BST
root = insert(root, arr[i]);
# Inorder Traversal to print nodes of Tree
inOrder(root);
# This code is contributed by AnkitRai01
C#
// C# program of above implementation
using System;
// Node declaration
public class Node
{
public int data;
public Node left;
public Node right;
public Node(int data)
{
this.data = data;
left = right = null;
}
}
class GFG{
// Function to insert data in BST
public static Node insert(Node root, int data)
{
if (root == null)
return new Node(data);
if (data < root.data)
root.left = insert(root.left, data);
if (data > root.data)
root.right = insert(root.right, data);
return root;
}
// InOrder function to display value of array
// in sorted order
public static void inOrder(Node root)
{
if (root == null)
return;
inOrder(root.left);
Console.Write(root.data + " ");
inOrder(root.right);
}
// Driver Code
static void Main()
{
int[] arr = { 1, 2, 3, 2, 5, 4, 4 };
// Finding size of array arr[]
int n = arr.Length;
Node root = null;
for(int i = 0; i < n; i++)
{
// Insert element of arr[] in BST
root = insert(root, arr[i]);
}
// Inorder Traversal to print nodes of Tree
inOrder(root);
}
}
// This code is contributed by divyeshrabadiya07
输出:
1 2 3 4 5
时间复杂度: O(N),其中N是给定数组的大小。
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