単独でリンクされたリストを考えると、タスクはリストの中央ノードを削除することです。
Pythonでタプルを並べ替える
- リストに偶数のノードが含まれている場合、2つの中央ノードがあります。この場合、2番目の中央ノードを削除します。
- リンクされたリストが1つのノードのみで構成されている場合、NULLを返します。
例：
入力： LinkedList：1-> 2-> 3-> 4-> 5
出力： 1-> 2-> 4-> 5
説明：
入力： LinkedList：2-> 4-> 6-> 7-> 5-> 1
出力： 2-> 4-> 6-> 5-> 1
説明：
入力： LinkedList：7
出力：
コンテンツの表
[ナイーブアプローチ] 2パストラバーサルの使用-O（n）時間とO（1）スペース
このアプローチの背後にある基本的なアイデアは、最初にリンクリスト全体を通過して、ノードの総数をカウントすることです。ノードの総数がわかったら、インデックスにある中間ノードの位置を計算できます n/2 （ここで、nはノードの総数です）。その後、リンクされたリストをもう一度通過しますが、今回は中央のノードの直前に停止します。そこに着いたら、中央のノードの前にノードの次のポインターを変更して、中央のノードをスキップしてノードを直接指しています。
以下は、上記のアプローチの実装です。
C++
// C++ program to delete middle of a linked list #include
// C program to delete middle of a linked list #include
// Java program to delete middle of a linked list class Node { int data; Node next; Node(int x) { data = x; next = null; } } public class GfG { // Function to delete middle node from linked list. public static Node deleteMid(Node head) { // Edge case: return null if there is only // one node. if (head.next == null) return null; int count = 0; Node p1 = head p2 = head; // First pass count the number of nodes // in the linked list using 'p1'. while (p1 != null) { count++; p1 = p1.next; } // Get the index of the node to be deleted. int middleIndex = count / 2; // Second pass let 'p2' move toward predecessor // of the middle node. for (int i = 0; i < middleIndex - 1; ++i) p2 = p2.next; // Delete the middle node and return 'head'. p2.next = p2.next.next; return head; } public static void printList(Node head) { Node temp = head; while (temp != null) { System.out.print(temp.data + ' -> '); temp = temp.next; } System.out.println('null'); } public static void main(String[] args) { // Create a static hardcoded linked list: // 1 -> 2 -> 3 -> 4 -> 5. Node head = new Node(1); head.next = new Node(2); head.next.next = new Node(3); head.next.next.next = new Node(4); head.next.next.next.next = new Node(5); System.out.print('Original Linked List: '); printList(head); // Delete the middle node. head = deleteMid(head); System.out.print ('Linked List after deleting the middle node: '); printList(head); } }
# Python3 program to delete middle of a linked list class Node: def __init__(self data): self.data = data self.next = None # Function to delete middle node from linked list. def deleteMid(head): # Edge case: return None if there is only # one node. if head.next is None: return None count = 0 p1 = head p2 = head # First pass count the number of nodes # in the linked list using 'p1'. while p1 is not None: count += 1 p1 = p1.next # Get the index of the node to be deleted. middleIndex = count // 2 # Second pass let 'p2' move toward the predecessor # of the middle node. for i in range(middleIndex - 1): p2 = p2.next # Delete the middle node and return 'head'. p2.next = p2.next.next return head def printList(head): temp = head while temp is not None: print(temp.data end=' -> ') temp = temp.next print('None') if __name__ == '__main__': # Create a static hardcoded linked list: # 1 -> 2 -> 3 -> 4 -> 5. head = Node(1) head.next = Node(2) head.next.next = Node(3) head.next.next.next = Node(4) head.next.next.next.next = Node(5) print('Original Linked List:' end=' ') printList(head) # Delete the middle node. head = deleteMid(head) print('Linked List after deleting the middle node:' end=' ') printList(head)
// C# program to delete middle of a linked list using System; class Node { public int data; public Node next; public Node(int x) { data = x; next = null; } } class GfG { // Function to delete middle node from linked list. static Node deleteMid(Node head) { // Edge case: return null if there is only // one node. if (head.next == null) return null; int count = 0; Node p1 = head p2 = head; // First pass count the number of nodes // in the linked list using 'p1'. while (p1 != null) { count++; p1 = p1.next; } // Get the index of the node to be deleted. int middleIndex = count / 2; // Second pass let 'p2' move toward the predecessor // of the middle node. for (int i = 0; i < middleIndex - 1; ++i) p2 = p2.next; // Delete the middle node and return 'head'. p2.next = p2.next.next; return head; } static void printList(Node head) { Node temp = head; while (temp != null) { Console.Write(temp.data + ' -> '); temp = temp.next; } Console.WriteLine('null'); } static void Main(string[] args) { // Create a static hardcoded linked list: // 1 -> 2 -> 3 -> 4 -> 5. Node head = new Node(1); head.next = new Node(2); head.next.next = new Node(3); head.next.next.next = new Node(4); head.next.next.next.next = new Node(5); Console.Write('Original Linked List: '); printList(head); // Delete the middle node. head = deleteMid(head); Console.Write ('Linked List after deleting the middle node: '); printList(head); } }
class Node { constructor(data) { this.data = data; this.next = null; } } // Function to delete middle node from linked list. function deleteMid(head) { // Edge case: return null if there is only // one node. if (head.next === null) return null; let count = 0; let p1 = head p2 = head; // First pass count the number of nodes // in the linked list using 'p1'. while (p1 !== null) { count++; p1 = p1.next; } // Get the index of the node to be deleted. let middleIndex = Math.floor(count / 2); // Second pass let 'p2' move toward the predecessor // of the middle node. for (let i = 0; i < middleIndex - 1; ++i) p2 = p2.next; // Delete the middle node and return 'head'. p2.next = p2.next.next; return head; } function printList(head) { let temp = head; while (temp !== null) { console.log(temp.data + ' -> '); temp = temp.next; } console.log('null'); } // Create a static hardcoded linked list: // 1 -> 2 -> 3 -> 4 -> 5. let head = new Node(1); head.next = new Node(2); head.next.next = new Node(3); head.next.next.next = new Node(4); head.next.next.next.next = new Node(5); console.log('Original Linked List: '); printList(head); // Delete the middle node. head = deleteMid(head); console.log('Linked List after deleting the middle node: '); printList(head);
出力
Original Linked List: 1 -> 2 -> 3 -> 4 -> 5 -> nullptr Linked List after deleting the middle node: 1 -> 2 -> 4 -> 5 -> nullptr
時間の複雑さ： の上）。 リンクリストの2つのトラバーサルが必要です
補助スペース： O（1）。 余分なスペースは必要ありません。
[予想されるアプローチ]ゆっくりと高速のポインターを備えたワンパストラバーサル-O（n）時間とO（1）スペース
上記のソリューションでは、リンクリストの2つのトラバーサルが必要です。中央のノードは、1つのトラバーサルを使用して削除できます。アイデアは、2つのポインターを使用することです slow_ptr そして fast_ptr 。高速ポインターは一度に2つのノードを移動し、遅いポインターが一度に1つのノードを移動します。高速ポインターがリストの最後に到達すると、遅いポインターが中央のノードに配置されます。次に、中央のノードの前に来るノードを接続する必要があります（ 前後 ）中央のノードの後に来るノードへ。これにより、ミドルノードを効果的にスキップして、リストから削除します。
以下は、上記のアプローチの実装です
C++// C++ program to delete middle of a linked list #include
// C program to delete middle of a linked list #include
// Java program to delete the middle of a linked list class Node { int data; Node next; Node(int x) { data = x; next = null; } } class GfG { // Function to delete middle node from linked list static Node deleteMid(Node head) { // If the list is empty return null if (head == null) return null; // If the list has only one node // delete it and return null if (head.next == null) { return null; } Node prev = null; Node slow_ptr = head; Node fast_ptr = head; // Move the fast pointer 2 nodes ahead // and the slow pointer 1 node ahead // until fast pointer reaches end of list while (fast_ptr != null && fast_ptr.next != null) { fast_ptr = fast_ptr.next.next; // Update prev to hold the previous // slow pointer value prev = slow_ptr; slow_ptr = slow_ptr.next; } // At this pointslow_ptr points to middle node // Bypass the middle node prev.next = slow_ptr.next; // Return the head of the modified list return head; } static void printList(Node head) { Node temp = head; while (temp != null) { System.out.print(temp.data + ' -> '); temp = temp.next; } System.out.println('NULL'); } public static void main(String[] args) { // Create a static hardcoded linked list: // 1 -> 2 -> 3 -> 4 -> 5 Node head = new Node(1); head.next = new Node(2); head.next.next = new Node(3); head.next.next.next = new Node(4); head.next.next.next.next = new Node(5); System.out.print('Original Linked List: '); printList(head); // Delete the middle node head = deleteMid(head); System.out.print ('Linked List after deleting the middle node: '); printList(head); } }
# Python program to delete the middle of a linked list class Node: def __init__(self data): self.data = data self.next = None # Function to delete middle node from linked list def deleteMid(head): # If the list is empty return None if head is None: return None # If the list has only one node # delete it and return None if head.next is None: return None prev = None slow_ptr = head fast_ptr = head # Move the fast pointer 2 nodes ahead # and the slow pointer 1 node ahead # until fast pointer reaches end of the list while fast_ptr is not None and fast_ptr.next is not None: fast_ptr = fast_ptr.next.next # Update prev to hold the previous # slow pointer value prev = slow_ptr slow_ptr = slow_ptr.next # At this point slow_ptr points to middle node # Bypass the middle node prev.next = slow_ptr.next # Return the head of the modified list return head def printList(head): temp = head while temp: print(temp.data end=' -> ') temp = temp.next print('NULL') if __name__ == '__main__': # Create a static hardcoded linked list: # 1 -> 2 -> 3 -> 4 -> 5 head = Node(1) head.next = Node(2) head.next.next = Node(3) head.next.next.next = Node(4) head.next.next.next.next = Node(5) print('Original Linked List: ' end='') printList(head) # Delete the middle node head = deleteMid(head) print('Linked List after deleting the middle node: ' end='') printList(head)
// C# program to delete middle of a linked list using System; class Node { public int data; public Node next; public Node(int x) { data = x; next = null; } } class GfG { // Function to delete middle node from linked list public static Node deleteMid(Node head) { // If the list is empty return null if (head == null) return null; // If the list has only one node // delete it and return null if (head.next == null) { return null; } Node prev = null; Node slow_ptr = head; Node fast_ptr = head; // Move the fast pointer 2 nodes ahead // and the slow pointer 1 node ahead // until fast pointer reaches end of the list while (fast_ptr != null && fast_ptr.next != null) { fast_ptr = fast_ptr.next.next; // Update prev to hold the previous // slow pointer value prev = slow_ptr; slow_ptr = slow_ptr.next; } // At this point slow_ptr points to middle node // Bypass the middle node prev.next = slow_ptr.next; // Return the head of the modified list return head; } // Function to print the linked list public static void printList(Node head) { Node temp = head; while (temp != null) { Console.Write(temp.data + ' -> '); temp = temp.next; } Console.WriteLine('NULL'); } public static void Main(string[] args) { // Create a static hardcoded linked list: // 1 -> 2 -> 3 -> 4 -> 5 Node head = new Node(1); head.next = new Node(2); head.next.next = new Node(3); head.next.next.next = new Node(4); head.next.next.next.next = new Node(5); Console.Write('Original Linked List: '); printList(head); // Delete the middle node head = deleteMid(head); Console.Write ('Linked List after deleting the middle node: '); printList(head); } }
// javascript program to delete middle of a linked list class Node { constructor(data) { this.data = data; this.next = null; } } // Function to delete the middle node from the linked list function deleteMid(head) { // If the list is empty return null if (head === null) { return null; } // If the list has only one node delete it and return // null if (head.next === null) { return null; } let prev = null; let slow_ptr = head; let fast_ptr = head; // Move the fast pointer 2 nodes ahead // and the slow pointer 1 node ahead // until the fast pointer reaches the end of the list while (fast_ptr !== null && fast_ptr.next !== null) { fast_ptr = fast_ptr.next.next; // Update prev to hold the previous slow pointer // value prev = slow_ptr; slow_ptr = slow_ptr.next; } // At this point slow_ptr points to the middle node // Bypass the middle node prev.next = slow_ptr.next; // Return the head of the modified list return head; } function printList(head) { let temp = head; while (temp !== null) { process.stdout.write(temp.data + ' -> '); temp = temp.next; } console.log('null'); } // Create a static hardcoded linked list: // 1 -> 2 -> 3 -> 4 -> 5 let head = new Node(1); head.next = new Node(2); head.next.next = new Node(3); head.next.next.next = new Node(4); head.next.next.next.next = new Node(5); process.stdout.write('Original Linked List: '); printList(head); // Delete the middle node head = deleteMid(head); process.stdout.write( 'Linked List after deleting the middle node: '); printList(head);
出力
Original Linked List: 1 -> 2 -> 3 -> 4 -> 5 -> NULL Linked List after deleting the middle node: 1 -> 2 -> 4 -> 5 -> NULL
時間の複雑さ： の上）。 リンクリストの1つのトラバーサルのみが必要です
補助スペース： O（1）。 余分なスペースが必要ないため。
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関連記事：
- リンクリストの中央を見つけます