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Given a linked list, remove the nth node from the end of list and return its head.
Given linked list: 1->2->3->4->5->null, and n = 2.
The minimum number of nodes in list is n.
O(n) time
SOLUTION:
笨方法就是走一遍,确定list的长度,然后再走一遍len-n的长度就行了,但是要求O(n)的话就要用到两个指针,第一个指针先走n步,第二个再走,当第一个指针已经到尾的时候,第二个指针跟尾自然相差n个,但是删除list的node的时候,要记录node前面的点,才能进行删除操作。
代码:
/** * Definition for ListNode. * public class ListNode { * int val; * ListNode next; * ListNode(int val) { * this.val = val; * this.next = null; * } * } */ public class Solution { /** * @param head: The first node of linked list. * @param n: An integer. * @return: The head of linked list. */ ListNode removeNthFromEnd(ListNode head, int n) { if (head == null || n < 1){ return head; } ListNode dummy = new ListNode(0); dummy.next = head; ListNode preDel = dummy; for (int i = 0; i < n; i++){ if (head != null){ head = head.next; } } while (head != null){ head = head.next; preDel = preDel.next; } preDel.next = preDel.next.next; return dummy.next; } }
[LintCode] Remove Nth Node From End of List
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原文地址:http://www.cnblogs.com/tritritri/p/4970821.html