<p>Given the <code>head</code> of a linked list, remove the <code>n<sup>th</sup></code> node from the end of the list and return its head.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2020/10/03/remove_ex1.jpg" style="width: 542px; height: 222px;" />
<pre>
<strong>Input:</strong> head = [1,2,3,4,5], n = 2
<strong>Output:</strong> [1,2,3,5]
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> head = [1], n = 1
<strong>Output:</strong> []
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> head = [1,2], n = 1
<strong>Output:</strong> [1]
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li>The number of nodes in the list is <code>sz</code>.</li>
	<li><code>1 &lt;= sz &lt;= 30</code></li>
	<li><code>0 &lt;= Node.val &lt;= 100</code></li>
	<li><code>1 &lt;= n &lt;= sz</code></li>
</ul>

<p>&nbsp;</p>
<p><strong>Follow up:</strong> Could you do this in one pass?</p>

/**
 * 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) {
        if(!head){
            return NULL;
        }
        if(head->next==NULL){
            return NULL;
        }
        ListNode*slow=head;
        ListNode*fast=head;
        while(n--){
            fast=fast->next;
            if(fast==NULL){
                return head->next;
            }
        }
        while(fast!=NULL && fast->next!=NULL){
            fast=fast->next;
            slow=slow->next;
        }
        ListNode*current=slow->next;
        ListNode*prev=slow;
        prev->next=current->next;
        current->next=NULL;
        delete current;
        return head;


    }
};
14600