LeetCode: Remove Nth Node From End of List 解题报告

Remove Nth Node From End of List

Total Accepted: 46720 Total Submissions: 168596My Submissions

Question Solution 

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.

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SOLUTION 1:

1.使用快慢指针，快指针先行移动N步。用慢指针指向要移除的Node的前一个Node.

2. 使用dummy node作为head的前缀节点，这样就算是删除head也能轻松handle啦！

主页君是不是很聪明呀？ :)

 /**
  * Definition for singly-linked list.
  * public class ListNode {
  *     int val;
  *     ListNode next;
  *     ListNode(int x) {
  *         val = x;
  *         next = null;
  *     }
  * }
  */
 public class Solution {
     public ListNode removeNthFromEnd(ListNode head, int n) {
         //
         ListNode dummy = new ListNode();
         dummy.next = head;

         ListNode slow = dummy;
         ListNode fast = dummy;

         // move fast N more than slow.
         while (n > ) {
             fast = fast.next;
             // Bug 1: FORGET THE N--;
             n--;
         }

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

         // Slow is the pre node of the node which we want to delete.
         slow.next = slow.next.next;

         return dummy.next;
     }
 }

GITHUB (国内用户可能无法连接):

https://github.com/yuzhangcmu/LeetCode/blob/251766ffb832f2278f43a05e194ca76584bf14ea/list/RemoveNthFromEnd.java