LeetCode OJ算法题（十九）：Remove Nth Node From End of List

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题目：

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.

解法：

题目很简单，就是链表的元素删除，思路有两种：

第一种，先遍历一边链表，得到链表的长度L，然后由L-n+1找到要删除的元素，这种时间复杂度为O（2n）

第二种，只遍历一遍链表，但遍历的同时，维持一个大小为n+1的数组，用来实时保存最近遍历到的n+1个元素，当遍历完成时，只需删除array[n]指向的元素即可。

当然需要注意，若array没有被填满，即n>=L时，需要单独处理下。

public class No19_RemoveNthNodeFromEndOfList {public static class ListNode {int val;ListNode next;ListNode(int x) {val = x;next = null;}}public static void main(String[] args){ListNode n1 = new ListNode(1);ListNode n2 = new ListNode(2);ListNode n3 = new ListNode(3);ListNode n4 = new ListNode(4);ListNode n5 = new ListNode(5);n1.next = n2;n2.next = n3;n3.next = n4;n4.next = n5;ListNode head = n1;//ListNode x = removeNthFromEnd_1(head, 2);ListNode x = removeNthFromEnd_2(head, 5);while(x!=null){System.out.println(x.val);x = x.next;}}public static ListNode removeNthFromEnd_1(ListNode head, int n) {int l = 0;ListNode p = head;while(p != null){p=p.next;l++;}if(n > l) return head;if(n == l) return head.next;int i = 1;p = head;while(i<l-n){p = p.next;i++;}ListNode tmp = p.next.next;p.next.next = null;p.next = tmp;        return head;    }public static ListNode removeNthFromEnd_2(ListNode head, int n) {ListNode[] nodeArr = new ListNode[n+1];ListNode p = head;while(p != null){for(int i=n;i>0;i--){nodeArr[i] = nodeArr[i-1];}nodeArr[0] = p;p = p.next;}if(nodeArr[n] == null) return head.next;ListNode tmp = nodeArr[n].next.next;nodeArr[n].next.next = null;nodeArr[n].next = tmp;        return head;    }}

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