19. Remove Nth Node From End of List

最新推荐文章于 2025-09-06 17:22:57 发布

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最新推荐文章于 2025-09-06 17:22:57 发布

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分类专栏： leetcode 文章标签： leetcode LinkedList two pointers

本文链接：https://blog.youkuaiyun.com/weixin_44476534/article/details/86622279

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本文探讨了在链表中删除倒数第N个节点的有效算法，提出了两种解决方案：一种是反转链表，然后进行删除再反转回原顺序；另一种是使用双指针技巧，一次遍历完成删除操作。这两种方法的时间复杂度均为O(N)，空间复杂度为O(1)。

Given a linked list, remove the n-th node from the end of list and return its head.

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.

Follow up:
Could you do this in one pass?

solution 1: reverse LinkedList

class Solution {
    public ListNode removeNthFromEnd(ListNode head, int n) {
        if (n < 1) {
            return head;
        }
        head = reverse(head);
        ListNode cur = head;
        if (n == 1) {
            return reverse(head.next);
        }
        while (n > 2) {
            cur = cur.next;
            n--;
        }
        cur.next = cur.next.next;
        return reverse(head);
        
    }
    private ListNode reverse(ListNode head) {
        if (head == null || head.next == null) {
            return head;
        }
        ListNode prev = null;
        while (head != null) {
            ListNode next = head.next;
            head.next = prev;
            prev = head;
            head = next;
        }
        return prev;
    }
}

time: O(N)
space: O(1)

solution 2: two pointers

make a pointer ahead n nodes of the second pointer, each eteration we move both pointer s one step until the first node meets null

public ListNode removeNthFromEnd(ListNode head, int n) {
    ListNode dummy = new ListNode(0);
    dummy.next = head;
    ListNode first = dummy;
    ListNode second = dummy;
    // Advances first pointer so that the gap between first and second is n nodes apart
    for (int i = 1; i <= n + 1; i++) {
        first = first.next;
    }
    // Move first to the end, maintaining the gap
    while (first != null) {
        first = first.next;
        second = second.next;
    }
    second.next = second.next.next;
    return dummy.next;
}