[LeetCode][Java] Remove Nth Node From End of List

题目：

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.

题意：

给定一个链表，去除距离尾节点第n个的节点，返回去除节点后的头结点。

例如：

   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.

注意：

给定的n都是有效的。

试着一次遍历就得出结果。

算法分析：

Use fast and slow pointers. The fast pointer is n steps ahead of the slow pointer. When the fast reaches the end, the slow pointer points at the previous element of the target
element.

利用快慢双指针。快指针在慢指针之前n步。当快指针到达链表的结尾时，慢指针正好指向了目标元素的头一个元素。

AC代码：

public class Solution
{
    public ListNode removeNthFromEnd(ListNode head, int n)
    {
        if(head == null)
            return null;

        ListNode fast = head;
        ListNode slow = head;

        for(int i=0; i<n; i++)
        {
            fast = fast.next;
        }

        //if remove the first node
        if(fast == null)
        {
            head = head.next;
            return head;
        }

        while(fast.next != null)
        {
            fast = fast.next;
            slow = slow.next;
        }

        slow.next = slow.next.next;

        return head;
    }
}

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