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

標籤：leetcode   java   remove nth node from   

題目：

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.

題意：

給定一個鏈表，去除距離尾節點第n個的節點，返回去除節點後的頭結點。

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：

給定的n都是有效。

試著一次遍曆就得出結果。

演算法分析：

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.

利用快慢雙指標。快指標在慢指標之前n步。當快指標到達鏈表的結尾時，慢指標正好指向了目標元素的頭一個元素。

AC代碼：

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