【算法】删除单链表的倒数第N个结点

Description

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.

这是Leetcode中的一道题，剑指offer中是查找到单链表的倒数第N个结点，题目类似。

解题思路

采用两个指针，first和slow，first先后移N个结点，然后slow和first同时向后移动，因为两个指针相差N个结点，所以当first.next==null，slow现在指的就是倒数第N个结点的前驱，所以直接利用slow.next = slow.next.next，删除这个节点。
代码如下：

/**
 * @ congrisheng
 * @ 2017-3-13
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
public ListNode removeNthFromEnd(ListNode head, int n){
    if(head == null || head.next == null){
        return null;
    }

    ListNode slow  = head;
    ListNode first = head; 

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

From《剑指offer》Page13