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.

题目意思是给你一个链表，让你去除倒数第n个节点。

需要注意的是：

1.Try to do this in one pass.

2.使用双指针。

即让current先跑n个节点，然后再让next和current一起跑，当current后面没有节点了，则代表next的当前节点是需要删除的节点。

package leetcode;

public class RemoveNthNodeFromEndofList {

public ListNode removeNthFromEnd(ListNodehead, int n) {

int size = 0;

ListNode current =head;

ListNode next = head;

if (head ==null) return null;

for (inti = 0; i<n;i++){

//System.out.println(current.val);

current =current.next;

}

if(current ==null)  

        {  

            head = head.next;  

            next = null;  

            return head;  

        } 

while (current.next !=null){

//System.out.println(current.val + " " + next.val);

current =current.next;

next =next.next;

}

//System.out.println(current.val + " " + next.val);

ListNode tmp = next.next.next;  

        next.next =tmp;

        

        return head;

    }

public staticvoid main(String[] args) {

// TODO Auto-generated method stub

ListNode head = new ListNode(1);

ListNode n1 = new ListNode(2);

ListNode n2 = new ListNode(3);

ListNode n3 = new ListNode(4);

ListNode end = new ListNode(5);

head.next =n1;

n1.next =n2;

n2.next =n3;

n3.next =end;

head = new RemoveNthNodeFromEndofList().removeNthFromEnd(head, 1);

while (head !=null){

System.out.print(head.val +" ");

head =head.next;

}

}

}