題目

? ? ? ? 這是道鏈表的簡單應用題目，刪除從結尾數第n個節點。

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.

代碼

/*** 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;} }

代碼下載：https://github.com/jimenbian/GarvinLeetCode

/********************************

* 本文來自博客 ?“李博Garvin“

* 轉載請標明出處:http://blog.csdn.net/buptgshengod

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