package com.freetymekiyan.algorithms.level.easy;
import com.freetymekiyan.algorithms.utils.Utils.ListNode;
/**
* Given a linked list, remove the nth node from the end of list and return its
* head.
* <p>
* For example,
* <p>
* Given linked list: 1->2->3->4->5, and n = 2.
* <p>
* After removing the second node from the end, the linked list becomes
* 1->2->3->5.
* <p>
* Note:
* Given n will always be valid.
* Try to do this in one pass.
* <p>
* Tags: Linked list, Two pointers
*/
class RemoveNthNodeFromEnd {
/**
* Dummy head and Runner's technique
*/
public ListNode removeNthFromEnd(ListNode head, int n) {
ListNode pre = new ListNode(0);
pre.next = head;
ListNode p1 = pre;
ListNode p2 = pre;
int i = 0;
while (i < n) {
p2 = p2.next;
i++;
}
while (p2.next != null) {
p1 = p1.next;
p2 = p2.next;
}
p1.next = p1.next.next;
return pre.next;
}
}