从给定数组创建链表
原文：https://www.geeksforgeeks.org/create-linked-list-from-a-given-array/
给定大小为N的数组arr[]。 任务是从给定数组创建链表。
示例：
Input : arr[]={1, 2, 3, 4, 5}
Output : 1->2->3->4->5
Input :arr[]={10, 11, 12, 13, 14}
Output : 10->11->12->13->14
简单方法：对于数组arr[]的每个元素，我们在链表中创建一个节点，并将其插入到末尾。
C++
#include <iostream>
using namespace std;
// Representation of a node
struct Node {
int data;
Node* next;
};
// Function to insert node
void insert(Node** root, int item)
{
Node* temp = new Node;
Node* ptr;
temp->data = item;
temp->next = NULL;
if (*root == NULL)
*root = temp;
else {
ptr = *root;
while (ptr->next != NULL)
ptr = ptr->next;
ptr->next = temp;
}
}
void display(Node* root)
{
while (root != NULL) {
cout << root->data << " ";
root = root->next;
}
}
Node *arrayToList(int arr[], int n)
{
Node *root = NULL;
for (int i = 0; i < n; i++)
insert(&root, arr[i]);
return root;
}
// Driver code
int main()
{
int arr[] = { 1, 2, 3, 4, 5 };
int n = sizeof(arr) / sizeof(arr[0]);
Node* root = arrayToList(arr, n);
display(root);
return 0;
}
Java
// Java implementation of the approach
import java.util.*;
class GFG
{
// Representation of a node
static class Node
{
int data;
Node next;
};
// Function to insert node
static Node insert(Node root, int item)
{
Node temp = new Node();
Node ptr;
temp.data = item;
temp.next = null;
if (root == null)
root = temp;
else
{
ptr = root;
while (ptr.next != null)
ptr = ptr.next;
ptr.next = temp;
}
return root;
}
static void display(Node root)
{
while (root != null)
{
System.out.print( root.data + " ");
root = root.next;
}
}
static Node arrayToList(int arr[], int n)
{
Node root = null;
for (int i = 0; i < n; i++)
root = insert(root, arr[i]);
return root;
}
// Driver code
public static void main(String args[])
{
int arr[] = { 1, 2, 3, 4, 5 };
int n = arr.length;
Node root = arrayToList(arr, n);
display(root);
}
}
// This code is contributed by Arnab Kundu
Python3
# Python3 implementation of the approach
import math
# Representation of a node
class Node:
def __init__(self, data):
self.data = data
self.next = None
# Function to insert node
def insert(root, item):
temp = Node(item)
if (root == None):
root = temp
else :
ptr = root
while (ptr.next != None):
ptr = ptr.next
ptr.next = temp
return root
def display(root):
while (root != None) :
print(root.data, end = " ")
root = root.next
def arrayToList(arr, n):
root = None
for i in range(0, n, 1):
root = insert(root, arr[i])
return root
# Driver code
if __name__=='__main__':
arr = [1, 2, 3, 4, 5]
n = len(arr)
root = arrayToList(arr, n)
display(root)
# This code is contributed by Srathore
C
// C# implementation of the above approach
using System;
class GFG
{
// Representation of a node
public class Node
{
public int data;
public Node next;
};
// Function to insert node
static Node insert(Node root, int item)
{
Node temp = new Node();
Node ptr;
temp.data = item;
temp.next = null;
if (root == null)
root = temp;
else
{
ptr = root;
while (ptr.next != null)
ptr = ptr.next;
ptr.next = temp;
}
return root;
}
static void display(Node root)
{
while (root != null)
{
Console.Write(root.data + " ");
root = root.next;
}
}
static Node arrayToList(int []arr, int n)
{
Node root = null;
for (int i = 0; i < n; i++)
root = insert(root, arr[i]);
return root;
}
// Driver code
public static void Main(String []args)
{
int []arr = { 1, 2, 3, 4, 5 };
int n = arr.Length;
Node root = arrayToList(arr, n);
display(root);
}
}
// This code is contributed by PrinciRaj1992
输出：
1 2 3 4 5
时间复杂度：O(n * n)
高效方法：我们从结尾遍历数组，并将每个元素插入列表的开头。
C++
#include <iostream>
using namespace std;
// Representation of a node
struct Node {
int data;
Node* next;
};
// Function to insert node
void insert(Node** root, int item)
{
Node* temp = new Node;
temp->data = item;
temp->next = *root;
*root = temp;
}
void display(Node* root)
{
while (root != NULL) {
cout << root->data << " ";
root = root->next;
}
}
Node *arrayToList(int arr[], int n)
{
Node *root = NULL;
for (int i = n-1; i >= 0 ; i--)
insert(&root, arr[i]);
return root;
}
// Driver code
int main()
{
int arr[] = { 1, 2, 3, 4, 5 };
int n = sizeof(arr) / sizeof(arr[0]);
Node* root = arrayToList(arr, n);
display(root);
return 0;
}
Java
// Java program to print level order traversal
// in spiral form using one queue and one stack.
import java.util.*;
class GFG
{
// Representation of a node
static class Node
{
int data;
Node next;
};
static Node root;
// Function to insert node
static Node insert(Node root, int item)
{
Node temp = new Node();
temp.data = item;
temp.next = root;
root = temp;
return root;
}
static void display(Node root)
{
while (root != null)
{
System.out.print(root.data + " ");
root = root.next;
}
}
static Node arrayToList(int arr[], int n)
{
root = null;
for (int i = n - 1; i >= 0 ; i--)
root = insert(root, arr[i]);
return root;
}
// Driver code
public static void main(String[] args)
{
int arr[] = { 1, 2, 3, 4, 5 };
int n = arr.length;
Node root = arrayToList(arr, n);
display(root);
}
}
// This code is contributed by Princi Singh
Python3
# Python3 program to prlevel order traversal
# in spiral form using one queue and one stack.
# Representation of a Node
class Node:
def __init__(self, data):
self.data = data
self.next = next
# Function to insert Node
def insert(root, item):
temp = Node(0)
temp.data = item
temp.next = root
root = temp
return root
def display(root):
while (root != None):
print(root.data, end=" ")
root = root.next
def arrayToList(arr, n):
root = None
for i in range(n - 1, -1, -1):
root = insert(root, arr[i])
return root
# Driver code
if __name__ == '__main__':
arr = [1, 2, 3, 4, 5];
n = len(arr)
root = arrayToList(arr, n);
display(root)
# This code is contributed by 29AjayKumar
C
// C# program to print level order traversal
// in spiral form using one queue and one stack.
using System;
class GFG
{
// Representation of a node
public class Node
{
public int data;
public Node next;
};
static Node root;
// Function to insert node
static Node insert(Node root, int item)
{
Node temp = new Node();
temp.data = item;
temp.next = root;
root = temp;
return root;
}
static void display(Node root)
{
while (root != null)
{
Console.Write(root.data + " ");
root = root.next;
}
}
static Node arrayToList(int []arr, int n)
{
root = null;
for (int i = n - 1; i >= 0 ; i--)
root = insert(root, arr[i]);
return root;
}
// Driver code
public static void Main(String[] args)
{
int []arr = { 1, 2, 3, 4, 5 };
int n = arr.Length;
Node root = arrayToList(arr, n);
display(root);
}
}
// This code is contributed by Rajput-Ji
输出：
1 2 3 4 5
时间复杂度：O(n)
替代有效解决方案是维护尾指针，从左到右遍历数组元素，在尾插入并在插入后更新尾。
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