Leecode #19 Remove Nth Node From End of List

一、 问题描述

Leecode第十九题，题目为：
Given a linked list, remove the n-th node from the end of list and return its head.

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.

Follow up:

Could you do this in one pass?
问题理解为：

给定一个链表，从链表末尾删除第n个节点并返回其头部。
例子:
给出链表:1->2->3->4->5,n = 2。
从末尾删除第二个节点后，链表变为1->2->3->5。
注意:
给定n总是有效的。
跟进:
你能一次做完吗?

二、算法思路
1、本题目应用快慢指针
2、

三、实现代码

/**
 * 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 a = head;
        ListNode b = head;

       
        for (int i = 0; i < n && a != null; i++) {
            a = a.next;
        }


        if (a == null) {
            head = head.next;
            return head;
        }

     
        while (a.next != null) {
            a = a.next;
            b = b.next;
        }

        b.next = b.next.next;
        return head;
    }
}