【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