LeetCode 019 Remove Nth Node From End of List

题目描述

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.

分析

题目中说n是合法的，就不用对n进行检查了。用标尺的思想，两个指针相距为n-1，后一个到表尾，则前一个到n了。

① 指针p、q指向链表头部；

② 移动q，使p和q差n-1；

③ 同时移动p和q，使q到表尾；

④ 删除p。

(p为second，q为first)

代码

[code]    public static ListNode removeNthFromEnd(ListNode head, int n) {

        if (head == null || head.next == null) {
            return null;
        }

        ListNode first = head;
        ListNode second = head;

        for (int i = 0; i < n; i++) {
            first = first.next;
            if (first == null) {
                return head.next;
            }
        }

        while (first.next != null) {
            first = first.next;
            second = second.next;
        }

        second.next = second.next.next;

        return head;
    }