# Leetcode 19.Remove Nth node from End of List (delete the last nth node) thinking and method of solving problems

Source: Internet
Author: User

Remove Nth Node from End of List

Given A linked list, remove the nth node from the end of the list and return its head.

For example,

Given linked list:1->2->3->4->5, and n = 2.

After removing the second node from the end, the linked list becomes 1->2->3->5.
Note:
Given n would always be valid.

Try to do the in one pass.

Idea: Delete the last nth node, because it is a single linked list, do not know the total number of nodes, it is first to traverse, count the total number of nodes, calculate the first number of positive numbers, and then delete.

The algorithm is simple and the code is as follows:

`/** * Definition for singly-linked list. * public class ListNode {*     int val; *     ListNode Next; *     listnode (int x) {val = x;}}} */public class Soluti On {public    ListNode removenthfromend (listnode head, int n) {        //delete the last nth        if (n = = 0) {            return head;        }        int nth = 0;//the penultimate nth        int count = 0;//Total number of nodes        listnode p = head;        Statistics Count        while (P! = null) {            p = p.next;            count++;        }        Calculates the value of the positive n, calculated from 0        n = count-n;        if (n = = 0) {//If 0, the header node, return to the head node next can            return head.next;        }        p = head;        Count to N-1, then make n-1.next = N.next = N-1.next.next        while (Nth < n-1) {            p = p.next;            nth++;        }        P.next = P.next.next;        return head;    }}`

Leetcode 19.Remove Nth node from End of List (delete the last nth node) ideas and methods of solving problems

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