Leetcode 19--Delete The penultimate node of a linked list

1. Topics

Given a linked list, deletes the penultimate node of the linked list and returns the head node of the linked list.

Example:
Given a linked list: 1->2->3->4->5, and n = 2.
When the second-to-last node is deleted, the list becomes 1->2->3->5.

Description
The given n guarantee is valid.

Advanced:
Can you try to use a scan to implement it?

2. Ideas

• Defines two pointers P1, p2, and just beginning both pointers point to the head node. If we want to delete the last nth node, we let P2 advance N-1 step , at this time the number of nodes behind the P2 is K, then we want to delete the node is located in the back of the P1 of the K node .

• As shown, suppose we want to delete the 3rd node, P2 forward 2 steps and point to the third node.

• There are 1 nodes behind the P2, and the node we want to delete is the 1th node at the back of P1-node 2.

• A special case is to delete the 4th node, the 1th node. At this point, we just need to have the head pointing to the 2nd node.

• In other cases, after the P2 pointer arrives at the specified position. Let's get the P2 pointer back one position , and then let P1 and P2 move backwards. When the P2 points to the last node, the P1 points to the previous one of the nodes to be deleted , allowing P1 to point to the 2nd node behind the P1 to delete the specified junction.

• The following is the process of removing the 1th node from the bottom.

• Of course, don't forget to release the memory of the deleted node .
• The code is as follows

/** * Definition for singly-linked list. * struct ListNode {* int val; * ListNode *next; * ListNode (int x): Val (x), Next (NULL) {} *};             */class Solution {public:listnode* removenthfromend (listnode* head, int n) {if (head = = NULL) {        return head;            } else {int i = 1;                        ListNode *p1 = head, *p2 = head, *temp = NULL;                while (i! = N)//P2 pointer forward n-1 step {i++;            P2 = p2->next;                } if (P2->next = = NULL)//Delete first node Case {temp = head; Head = head->next;                            Head points to 2nd node} else {P2 = p2->next;//p2 pointer previously in 1 steps                    while (P2->next)//P1, p2 pointer synchronously moves backwards {P2 = p2->next;                P1 = p1->next;    } temp = p1->next;            P1->next = p1->next->next; P1 points to the 2nd node behind it, delete (temp);                Release node memory} return head; }        };

For more highlights, please follow "seniusen"!

Leetcode 19--Delete The penultimate node of a linked list