019，Remove Nth Node From End of

http://blog.csdn.net/DERRANTCM/article/details/46997239

https://leetcode.com/problems/remove-nth-node-from-end-of-list/description/

原题

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.

题目大意

删除单链表的倒数第Ｎ个结点，注意：输入的Ｎ都是合法，在一次遍历中完成操作。

解题思路

先让一个指针走找到第N个节点，然后再让一个指针指向头结点，然后两具指针一起走，直到前一个指针直到了末尾，后一个指针就是倒数第N+1个结点，删除倒数第N个结点就可以了。

public class Solution {
    public ListNode removeNthFromEnd(ListNode head, int n) {
        ListNode pa = head;
        ListNode pb = head;

        // 找到第n个结点
        for (int i = 0; i < n && pa != null; i++) {
            pa = pa.next;
        }


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

        // pb与pa相差n-1个结点
        // 当pa.next为null，pb在倒数第n+1个位置
        while (pa.next != null) {
            pa = pa.next;
            pb = pb.next;
        }

        pb.next = pb.next.next; //pb.next指向下下个节点
        return head;
    }
}

个人理解：在红色部分思考半天，原来是把这个倒数的第N个直接删除，直接返回链表，醉了