LeetCode | Remove Nth Node From End of List

题目

Given a linked list, remove the n^th 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.

有可能删掉头部节点，因此用个哨兵可以简化思路和代码。

题目已经保证了n的可用，因此可以省去很多判断了。

代码

public class RemoveNthNodeFromEndOfList {
	public ListNode removeNthFromEnd(ListNode head, int n) {
		ListNode dummy = new ListNode(0);
		dummy.next = head;
		ListNode p = dummy;
		for (int i = 0; i < n; ++i) {
			p = p.next;
		}
		ListNode q = dummy;
		while (p.next != null) {
			p = p.next;
			q = q.next;
		}
		q.next = q.next.next;
		return dummy.next;
	}
}

LeetCode | Remove Nth Node From End of List,布布扣,bubuko.com

LeetCode | Remove Nth Node From End of List

原文：http://blog.csdn.net/perfect8886/article/details/23040515