LeetCode算法入门- Remove Nth Node From End of List -day17

1. 题目解释：

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.

题意就是删除倒数第n个节点

2. 题意分析：使用两个指针，一个快，一个慢，快的先走n步，当快的走到尾部null的时候，慢指针所指在就是要删除在那个节点。

3. Java实现：

/*** Definition for singly-linked list.* public class ListNode {*     int val;*     ListNode next;*     ListNode(int x) { val = x; }* }*/
class Solution {public ListNode removeNthFromEnd(ListNode head, int n) {if(head == null){return null;}ListNode fastNode = head;ListNode slowNode = head;//先提前让快指针走n步while(n-- > 0){fastNode = fastNode.next;}//记得加这一步在判断，什么情况下会为null（1->2  n = 2）,翻译过来就是要删掉头部//所以就是要特殊考虑删掉头部的情况if(fastNode == null){head = head.next;return head;}//算法的核心部分while(fastNode.next != null){fastNode = fastNode.next;slowNode = slowNode.next;}//删掉倒数第n个节点的操作，将slowNode节点直接指向slowNode的next。next节点，即可删掉slowNode.next = slowNode.next.next;return head;}
}