LeetCode

题目：

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.

思路：

三个指针p1,p2,p3，让第一指针p1走n - 1步停下来，然后p1和p2一起往前走，同时保持一个指针指向p2的前一个节点。

package list;

public class RemoveNthNodeFromEndOfList {

    public ListNode removeNthFromEnd(ListNode head, int n) {
        ListNode p1 = head;
        ListNode p2 = head;
        ListNode prevp2 = new ListNode(0);
        prevp2.next = p2;
        head = prevp2;
        while (--n > 0) {
            p1 = p1.next;
        }
        
        while (p1.next != null) {
            p1 = p1.next;
            prevp2 = p2;
            p2 = p2.next;
        }
        
        prevp2.next = p2.next;
        return head.next;
    }
    
    public static void main(String[] args) {
        // TODO Auto-generated method stub

    }

}