## [19\. Remove Nth Node From End of List](https://leetcode.com/problems/remove-nth-node-from-end-of-list/)

Given the `head` of a linked list, remove the `nth` node from the end of the list and return its head.

**Example 1:**


**Input:** head = \[1,2,3,4,5\], n = 2
**Output:** \[1,2,3,5\]

**Example 2:**

**Input:** head = \[1\], n = 1
**Output:** \[\]

**Example 3:**

**Input:** head = \[1,2\], n = 1
**Output:** \[1\]

**Constraints:**

- The number of nodes in the list is `sz`.
- `1 <= sz <= 30`
- `0 <= Node.val <= 100`
- `1 <= n <= sz`

**Follow up:** Could you do this in one pass?

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode() : val(0), next(nullptr) {}
 *     ListNode(int x) : val(x), next(nullptr) {}
 *     ListNode(int x, ListNode *next) : val(x), next(next) {}
 * };
 */
class Solution {
public:
    ListNode* removeNthFromEnd(ListNode* head, int n) {
        ListNode* slow = head;
        ListNode* fast = head;
        for(int i = 0; i < n; i++) {
            fast = fast->next;
        }

        if(!fast) return head->next;

        while(fast && fast->next) {
            slow = slow->next;
            fast = fast->next;
        }
        slow->next = slow->next->next;
        return head;
    }
};

:::success
- 時間複雜度：$O(N)$
- 空間複雜度：$O(1)$
:::