19. Remove Nth Node From End of List

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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.
题意：删除链表的倒数第n个数字
方法一：先遍历一遍链表，求出链表的长度，再删除length-n+1的数字

public class Solution {    public class ListNode {          int val;          ListNode next;          ListNode(int x) { val = x; }      }     public ListNode removeNthFromEnd(ListNode head, int n) {            if (head == null) {                  return null;              }              ListNode temp = head;              int count = 1;              while (temp.next != null) {                  count++;                  temp = temp.next;              }              if ((count - n + 1) == 1) {                  return head.next;              }              temp = head;              int i = 0;              while (temp != null) {                  i++;                  if (i == count - n) {                      temp.next = temp.next.next;                  }                  temp = temp.next;              }              return head;          }}

方法二：快慢指针，快指针和慢指针之间保持n的距离，当快指针遍历到最后一个数字时，慢指针正好在倒数第n个，只需要删除慢指针指向的数字即可

 if(head == null)        return null;    ListNode fast = head;    ListNode slow = head;    for(int i=0; i<n; i++){        fast = fast.next;    }    //if remove the first node    if(fast == null){        head = head.next;        return head;    }    while(fast.next != null){        fast = fast.next;        slow = slow.next;    }    slow.next = slow.next.next;    return head;

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