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.

 

 这道题要注意的Corner Case是：如果n比这个LinkedList的size大，那么就需要直接返回Head，如果相等，就把head删掉，返回head.next；基本做的方法呢，还是Runner Technique. 由于有可能head会被删掉，所以最好还是使用一个dummy node，dummy.next = head。

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
class Solution {
    public ListNode removeNthFromEnd(ListNode head, int n) {
        ListNode dummy = new ListNode(-1);
        dummy.next = head;
        ListNode p1 = dummy, p2 = dummy;
        while (p2.next != null) {
            if (n > 0) {
                n --;
            }
            else p1 = p1.next;
            p2 = p2.next;
        }
        if (n == 0) p1.next = p1.next.next;
        return dummy.next;
    }
}