删除链表的最后一个节点

给定一个链表，任务是移除链表的最后一个节点并更新链表的头指针。

例子：

Input: 1 -> 2 -> 3 -> 4 -> 5 -> NULL
Output: 1 -> 2 -> 3 -> 4 -> NULL

Explanation: The last node of the linked list
is 5, so 5 is deleted.

Input: 2 -> 4 -> 6 -> 8 -> 33 -> 67 -> NULL
Output: 2 -> 4 -> 6 -> 8 -> 33 -> NULL

Explanation: The last node of the linked list
is 67, so 67 is deleted. 

方法：要删除链表的最后一个节点，找到倒数第二个节点并使该节点的next指针为空。

算法：

1. 如果第一个节点为空或只有一个节点，则返回空。

if headNode == null then return null
if headNode.nextNode == null then free 
head and return null

2.创建一个额外的空间secondLast ，并遍历链表直到倒数第二个节点。

while secondLast.nextNode.nextNode != null 
      secondLast = secondLast.nextNode 

3.删除最后一个节点，即倒数第二个节点delete( secondLast.nextNode )的下一个节点，设置倒数第二个节点的值为null。

执行：

C++

// CPP program to remove last node of
// linked list.
#include 
using namespace std;
 
/* Link list node */
struct Node {
    int data;
    struct Node* next;
};
 
/* Function to remove the last node 
   of the linked list */
Node* removeLastNode(struct Node* head)
{
    if (head == NULL)
        return NULL;
 
    if (head->next == NULL) {
        delete head;
        return NULL;
    }
 
    // Find the second last node
    Node* second_last = head;
    while (second_last->next->next != NULL)
        second_last = second_last->next;
 
    // Delete last node
    delete (second_last->next);
 
    // Change next of second last
    second_last->next = NULL;
 
    return head;
}
 
// Function to push node at head
void push(struct Node** head_ref, int new_data)
{
    struct Node* new_node = new Node;
    new_node->data = new_data;
    new_node->next = (*head_ref);
    (*head_ref) = new_node;
}
 
// Driver code
int main()
{
    /* Start with the empty list */
    Node* head = NULL;
 
    /* Use push() function to construct  
       the below list 8 -> 23 -> 11 -> 29 -> 12 */
    push(&head, 12);
    push(&head, 29);
    push(&head, 11);
    push(&head, 23);
    push(&head, 8);
 
    head = removeLastNode(head);
    for (Node* temp = head; temp != NULL; temp = temp->next)
        cout << temp->data << " ";
 
    return 0;
}

Java

// Java program to remove last node of
// linked list.
class GFG {
 
    // Link list node /
    static class Node {
        int data;
        Node next;
    };
 
    // Function to remove the last node
    // of the linked list /
    static Node removeLastNode(Node head)
    {
        if (head == null)
            return null;
 
        if (head.next == null) {
            return null;
        }
 
        // Find the second last node
        Node second_last = head;
        while (second_last.next.next != null)
            second_last = second_last.next;
 
        // Change next of second last
        second_last.next = null;
 
        return head;
    }
 
    // Function to push node at head
    static Node push(Node head_ref, int new_data)
    {
        Node new_node = new Node();
        new_node.data = new_data;
        new_node.next = (head_ref);
        (head_ref) = new_node;
        return head_ref;
    }
 
    // Driver code
    public static void main(String args[])
    {
        // Start with the empty list /
        Node head = null;
 
        // Use push() function to con
        // the below list 8 . 23 . 11 . 29 . 12 /
        head = push(head, 12);
        head = push(head, 29);
        head = push(head, 11);
        head = push(head, 23);
        head = push(head, 8);
 
        head = removeLastNode(head);
        for (Node temp = head; temp != null; temp = temp.next)
            System.out.print(temp.data + " ");
    }
}
 
// This code is contributed by Arnab Kundu

Python3

# Python3 program to remove the last node of
# linked list.
import sys
import math
 
# Link list node
class Node:
    def __init__(self, data):
        self.data = data
        self.next = None
 
# Function to push node at head
def push(head, data):
    if not head:
        return Node(data)
    temp = Node(data)
    temp.next = head
    head = temp
    return head
 
# Function to remove the last node
# of the linked list
def removeLastNode(head):
    if head == None:
        return None
    if head.next == None:
        head = None
        return None
    second_last = head
    while(second_last.next.next):
        second_last = second_last.next
    second_last.next = None
    return head
 
# Driver code
if __name__=='__main__':
 
    # Start with the empty list
    head = None
    # Use push() function to con
    # the below list 8 . 23 . 11 . 29 . 12
    head = push(head, 12)
    head = push(head, 29)
    head = push(head, 11)
    head = push(head, 23)
    head = push(head, 8)
 
    head = removeLastNode(head)
    while(head):
        print("{} ".format(head.data), end ="")
        head = head.next
 
# This code is contributed by Vikash kumar 37

C#

// C# program to remove last node of
// linked list.
using System;
 
class GFG {
 
    // Link list node /
    public class Node {
        public int data;
        public Node next;
    };
 
    // Function to remove the last node
    // of the linked list /
    static Node removeLastNode(Node head)
    {
        if (head == null)
            return null;
 
        if (head.next == null) {
            return null;
        }
 
        // Find the second last node
        Node second_last = head;
        while (second_last.next.next != null)
            second_last = second_last.next;
 
        // Change next of second last
        second_last.next = null;
 
        return head;
    }
 
    // Function to push node at head
    static Node push(Node head_ref, int new_data)
    {
        Node new_node = new Node();
        new_node.data = new_data;
        new_node.next = (head_ref);
        (head_ref) = new_node;
        return head_ref;
    }
 
    // Driver code
    public static void Main(String[] args)
    {
        // Start with the empty list /
        Node head = null;
 
        // Use push() function to con
        // the below list 8 . 23 . 11 . 29 . 12 /
        head = push(head, 12);
        head = push(head, 29);
        head = push(head, 11);
        head = push(head, 23);
        head = push(head, 8);
 
        head = removeLastNode(head);
        for (Node temp = head; temp != null; temp = temp.next)
            Console.Write(temp.data + " ");
    }
}
 
/* This code contributed by PrinciRaj1992 */

Javascript

// javascript program to remove last node of
// linked list.   // Link list node /
    class Node {
        constructor() {
            this.data = 0;
            this.next = null;
        }
    }
    // Function to remove the last node
    // of the linked list /
    function removeLastNode(head)
    {
        if (head == null)
            return null;
 
        if (head.next == null) {
            return null;
        }
 
        // Find the second last node
        var second_last = head;
        while (second_last.next.next != null)
            second_last = second_last.next;
 
        // Change next of second last
        second_last.next = null;
 
        return head;
    }
 
    // Function to push node at head
    function push(head_ref , new_data)
    {
        var new_node = new Node();
        new_node.data = new_data;
        new_node.next = (head_ref);
        (head_ref) = new_node;
        return head_ref;
    }
 
    // Driver code
     
     
        // Start with the empty list /
        var head = null;
 
        // Use push() function to con
        // the below list 8 . 23 . 11 . 29 . 12 /
        head = push(head, 12);
        head = push(head, 29);
        head = push(head, 11);
        head = push(head, 23);
        head = push(head, 8);
 
        head = removeLastNode(head);
        for (temp = head; temp != null; temp = temp.next)
            document.write(temp.data + " ");
 
// This code contributed by Rajput-Ji

输出：

8 23 11 29

复杂性分析：

• 时间复杂度： O(n)。
  该算法涉及到链表的遍历直到其结束，因此所需的时间复杂度为O(n) 。
• 空间复杂度： O(1)。
  不需要额外的空间，因此空间复杂度是恒定的。