LeetCode 19. Remove Nth Node From End of List

原题链接在这里：https://leetcode.com/problems/remove-nth-node-from-end-of-list/

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.

题解：

Method 1:算出总长度, 再减去n, 即为要从头多动的点。但要新要求, only one pass.

Method 2:两个指针，一个runner, 一个walker, runner先走n步，随后runner和walker一起走，直到runner指为空。

Note:1. 都是找到要去掉点的前一个点记为mark, 再通过mark.next = mark.next.next去掉对应应去掉的点.

2. 注意去掉list头点的情况，所以要建立一个dummy head. e.g. 1->2, n = 2.

Time Complexity: O(n), n是list长度. Space: O(1).

AC Java: 

/**
 * 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) {
        /*Method 1
        if(head == null)
            return head;
        int len = 0;
        ListNode temp = head;
        while(temp != null){
            temp = temp.next;
            len++;
        }
        if(len < n){
            return head;
        }
        int moveNum = len - n;
        ListNode dunmy = new ListNode(0);
        dunmy.next = head;
        temp = dunmy;
        while(moveNum > 0){
            temp = temp.next;
            moveNum--;
        }
        temp.next = temp.next.next;
        return dunmy.next;
        */
        
        //Method 2
        if(head == null || n == 0)
            return head;

        ListNode dummy = new ListNode(0);
        dummy.next = head;
        ListNode walker = dummy;
        ListNode runner = dummy;
        
        while(n>0 && runner.next != null){
            runner = runner.next;
            n--;
        }
        
        while(runner.next != null){
            walker = walker.next;
            runner = runner.next;
        }
        
        walker.next = walker.next.next;
        return dummy.next;
    }
}