【算法】删除单链表的倒数第N个结点

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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

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