LeetCode 19 删除链表倒数第N个节点

// 方法一: 数有多少个节点,得到正着数的编号,共需两遍遍历
// 方法二: 双指针,first先走N步,然后second随着first一起走,知道first走到末尾,
//        删掉second后面的节点即可,只需一遍遍历
class Solution {

    // 方法一
    public ListNode removeNthFromEnd(ListNode head, int n) {

        if (head == null)
            return new ListNode(1); // 防止为空,但应该没有这样的样例

        if (n == 0)
            return head;
        int length = 0;
        ListNode p = head;
        while (p != null) {
            p = p.next;
            length++;
        }
        if (n > length)
            return head;
        if (n == length)
            return head.next;

        p = head;
        length = length - n;
        while (length > 1) {
            p = p.next;
            length--;
        }
        if (n == 1) {
            p.next = null;
            return head;
        }

        p.next = p.next.next;
        return head;
    }

    // 方法二 

    public ListNode removeNthFromEnd(ListNode head, int n) {
        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;
    }
}