LeetCode - Remove Nth Node From End of List

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.

思路：

三个指针p1,p2,p3，让第一指针p1走n - 1步停下来，然后p1和p2一起往前走，同时保持一个指针指向p2的前一个节点。

package list;public class RemoveNthNodeFromEndOfList {public ListNode removeNthFromEnd(ListNode head, int n) {ListNode p1 = head;ListNode p2 = head;ListNode prevp2 = new ListNode(0);prevp2.next = p2;head = prevp2;while (--n > 0) {p1 = p1.next;}while (p1.next != null) {p1 = p1.next;prevp2 = p2;p2 = p2.next;}prevp2.next = p2.next;return head.next;}public static void main(String[] args) {// TODO Auto-generated method stub
}}

转载于:https://www.cnblogs.com/null00/p/5060233.html

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