Remove Nth Node From End of List

最新推荐文章于 2024-01-12 15:01:11 发布

Paul_ZJ

最新推荐文章于 2024-01-12 15:01:11 发布

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分类专栏： LeetCode——坚持每天一题

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本文介绍了一种高效移除链表中倒数第N个节点的方法，并提供了一个Java实现示例。该方法仅遍历一次链表即可完成节点移除操作。

Remove Nth Node From End of List

Given a linked list, remove the n^th 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.

Java代码：

/**
 * 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) {
        int len =1;
        ListNode tmp =head;
        ListNode tmp_2;
 
        while(tmp.next != null){
        	tmp = tmp.next;
        	len++;
        }
        if(n==len){
        	tmp_2 = head.next;
        	head.next=null;
        	return tmp_2;
        }
        len=len-n-1;
        if(len<0)return null;
        tmp =head;
        while(len>0){
        	tmp=tmp.next;
        	len--;
        }
        tmp_2 = tmp.next;
        tmp.next=tmp.next.next;
        tmp_2.next=null;
        return head;
    }
}