LeetCode : Remove Nth Node From End of List [java]

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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：快指针首先移动N步，之后慢指针和快指针同时移动，当快指针走到末尾时，慢指针指向的正好是要删除的元素。

/** * Definition for singly-linked list. * public class ListNode { *     int val; *     ListNode next; *     ListNode(int x) { val = x; } * } */public class Solution {public ListNode removeNthFromEnd(ListNode head, int n) {if (head == null) {return null;}ListNode p1 = head; // quick pointerListNode p2 = head; // slow pointerListNode p3 = p2; // before slow pointerListNode result = p2;for (int i = 0; i < n - 1; i++) {p1 = p1.next;}if (p1.next == null) {return head.next;}while (p1.next != null) {p1 = p1.next;p3 = p2;p2 = p2.next;}p3.next = p2.next;return result;}}

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