Remove Nth Node from End of List

https://leetcode.com/problems/remove-nth-node-from-end-of-list/

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,

   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.

Title: Delete the last nth node of the linked list.

How does the countdown work? Using dummy nodes and two pointers P1 and P2,p1 first n steps, then P1 and P2 go together, when P1.next is empty, when the entire list is gone, P2.next is the reciprocal nth node to be deleted.

(Set the list of M nodes, you need to find the penultimate nth, then P2 need to go forward and through the M-n node, the next is the desired nth node, the positive sequence of m-n step how to control, is through the P1 this pointer)

1 #Definition for singly-linked list.2 #class ListNode:3 #def __init__ (self, x):4 #self.val = x5 #Self.next = None6 7 classSolution:8     #@param {ListNode} head9     #@param {integer} nTen     #@return {ListNode} One     defremoventhfromend (self, head, N): Adummy=listnode (0) -dummy.next=Head -P1=p2=dummy#p1,p2 simultaneous dummy departure the          forIinchRange (N):#P1 go first n step -p1=P1.next -          whileP1.next:#p1,p2 go at the same time, until P1 walk through the list -p1=P1.next +P2=P2.next -P2.next=p2.next.next#Delete the p2.next at this time +         returnDummy.next

Remove Nth Node from End of List