Leetcode-19. Remove Nth Node From End of List

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前言：为了后续的实习面试，开始疯狂刷题，非常欢迎志同道合的朋友一起交流。因为时间比较紧张，目前的规划是先过一遍，写出能想到的最优算法，第二遍再考虑最优或者较优的方法。如有错误欢迎指正。博主首发CSDN，mcf171专栏。

博客链接：mcf171的博客

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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.

Note:
Given n will always be valid.

Try to do this in one pass.

首先想到的就是先遍历一遍，确定深度，然后在遍历一遍把需要修改的地方改了。Your runtime beats 27.76% of java submissions.

/** * 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) {        int level = 0;ListNode search = head;while(search.next != null){search = search.next;level ++;}if( (level - n + 1) == 0) return head.next;level = level - n + 1;search = head;ListNode lastNode = search;while(level > 0){lastNode = search;search = search.next;level --;}lastNode.next = search.next;return head;    }}

看了几个解决方案，其实都是伪一次访问，我想了想，只有增加或者改变数据结构，否则不可能只浏览一次，因此加入Hash的数据结构，但是奇怪的是为什么时间效率还变低了，应该变快了才对。Your runtime beats 5.06% of java submissions.

public class Solution {    public ListNode removeNthFromEnd(ListNode head, int n) {        Map<Integer,ListNode> nodes = new HashMap<Integer,ListNode>();ListNode search = head;int deep = 0;nodes.put(deep,search);while(search.next != null){nodes.put(deep,search);deep++;search = search.next;}int delNumber = deep - n + 1;if(delNumber == 0 ) return head.next;else{ListNode lastNode = nodes.get(delNumber - 1);lastNode.next = lastNode.next.next;}return head;    }}

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