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【leetcode刷题笔记】Remove Nth Node From End of List

2024-07-13 02:22:38   219人阅读

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.

 

题解：快慢指针的思想，就像找到链表中点或者判断链表是否有环一样，是很经典的思想。

设置fast指针比slow指针多走n步，然后slow和fast一起前进，当fast指向null的时候，slow就指向从后往前数的第n个元素了。

但是要删除某个节点，最好的是知道它前面的节点，所以slow实际比fast慢n+1步。

代码如下：

 1 /** 2  * Definition for singly-linked list. 3  * public class ListNode { 4  *     int val; 5  *     ListNode next; 6  *     ListNode(int x) { 7  *         val = x; 8  *         next = null; 9  *     }10  * }11  */12 public class Solution {13     public ListNode removeNthFromEnd(ListNode head, int n) {14         if(head == null)15             return null;16         17         ListNode slow = new ListNode(0);18         ListNode answer = slow;19         slow.next = head;20         ListNode fast = head;21         for(int i = 0;i < n;i++){22             if(fast == null)23                 return null;24             fast = fast.next;25         }26         while(fast != null){27             slow = slow.next;28             fast = fast.next;29         }30         31         //delete what slow.next32         slow.next = slow.next.next;33         return answer.next;34     }35 }


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