leetcode : Remove Nth Node From End of List

Given a linked list, remove the nth node from the end of 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 will always be valid.
Try to do this in one pass.

tag: two pointers, dummy node.  尽量少的指针完成。

链表基础： reverse linked list 

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
public class Solution {
    public ListNode removeNthFromEnd(ListNode head, int n) {
     
        ListNode dummy = new ListNode(0);
        dummy.next = head;
        ListNode preN = dummy;
        for(int i = 0; i < n; i++) {
            if(head == null) {
                return null;
            }
            head = head.next;
        }
        
        while(head != null) {
            head = head.next;
            preN = preN.next;
        }
        preN.next = preN.next.next;
        return dummy.next;
    }
}