19. Remove Nth Node From End of List

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#leetcode #Remove Nth Node From

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本文介绍了一种高效算法，用于从链表中删除倒数第N个节点，仅通过一次遍历实现。提供了详细的解决方案及Java代码实现。

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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.

Solution:

Tips:

two pointers, one pointer goes first, and then the other goes after the first pointer's N steps.

Java Code:

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
public class Solution {
    public ListNode removeNthFromEnd(ListNode head, int n) {
        if (head == null) {
            return head;
        }
        
        ListNode tmpNode = head;
        ListNode tmpNodeNth = head;
        while ((n--) > 0) {
            tmpNodeNth = tmpNodeNth.next;
        }
        if (tmpNodeNth == null) {
            return head.next;
        }
        
        tmpNodeNth = tmpNodeNth.next;
        while (tmpNodeNth != null) {
            tmpNodeNth = tmpNodeNth.next;
            tmpNode = tmpNode.next;
        }
        tmpNode.next = tmpNode.next.next;
        
        return head;
    }
}