题目

这是道链表的简单应用题目，删除从结尾数第n个节点。

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.

代码

/*** Definition for singly-linked list.* public class ListNode {*     int val;*     ListNode next;*     ListNode(int x) {*         val = x;*         next = null;*     }* }*/public class Solution {public ListNode removeNthFromEnd(ListNode head, int n) {ListNode faster = head;ListNode slower = head;while (n > 0 && faster != null) {faster = faster.next;n--;}// Check if has only nodeif (faster == null) return head.next; while (faster.next != null) {faster = faster.next;slower = slower.next;}// Remove slower.next which is the nth form the endslower.next = slower.next.next;return head;}
}

代码下载：https://github.com/jimenbian/GarvinLeetCode

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【LeetCode从零单排】No19.RemoveNthNodeFromEndofList

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