LeetCode每日一题：remove nth node fro

问题描述

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.

问题分析

这道题只能遍历一次，很自然的想到了快慢指针，直接连接待删除点前后两个节点即可。java不用自己释放内存，让jvm处理。

代码实现

public ListNode removeNthFromEnd(ListNode head, int n) {
        if (head == null) return head;
        ListNode preHead = new ListNode(0);
        preHead.next = head;
        ListNode slow, fast;
        slow = fast = head;
        for (int i = 0; i < n; i++) {
            fast = fast.next;
        }
        if (fast == null) return slow.next;
        while (fast.next != null) {
            fast = fast.next;
            slow = slow.next;
        }
        slow.next = slow.next.next;
        return preHead.next;
    }