[Problem]
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.
[Analysis]
用固定距离的double pointer解决。
[Solution]
public class Solution { public ListNode removeNthFromEnd(ListNode head, int n) { int i = 1; ListNode node = head; ListNode pre = new ListNode(0); pre.next = head; while (node.next != null) { node = node.next; if (++i > n) { pre = pre.next; } } if (pre.next == head) { return head.next; } pre.next = pre.next.next; return head; } }
LeetCode #19 Remove Nth Node From End of List (E)
原文：http://www.cnblogs.com/zhangqieyi/p/4865528.html