<p>Given the <code>head</code> of a linked list, reverse the nodes of the list <code>k</code> at a time, and return <em>the modified list</em>.</p>

<p><code>k</code> is a positive integer and is less than or equal to the length of the linked list. If the number of nodes is not a multiple of <code>k</code> then left-out nodes, in the end, should remain as it is.</p>

<p>You may not alter the values in the list&#39;s nodes, only nodes themselves may be changed.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2020/10/03/reverse_ex1.jpg" style="width: 542px; height: 222px;" />
<pre>
<strong>Input:</strong> head = [1,2,3,4,5], k = 2
<strong>Output:</strong> [2,1,4,3,5]
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2020/10/03/reverse_ex2.jpg" style="width: 542px; height: 222px;" />
<pre>
<strong>Input:</strong> head = [1,2,3,4,5], k = 3
<strong>Output:</strong> [3,2,1,4,5]
</pre>

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

<ul>
	<li>The number of nodes in the list is <code>n</code>.</li>
	<li><code>1 &lt;= k &lt;= n &lt;= 5000</code></li>
	<li><code>0 &lt;= Node.val &lt;= 1000</code></li>
</ul>

<p>&nbsp;</p>
<p><strong>Follow-up:</strong> Can you solve the problem in <code>O(1)</code> extra memory space?</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:
    void reverse(ListNode* fs, ListNode* prev, int k){
        for(int i(0);i<k;++i){
            ListNode* front=fs->next;
            fs->next=prev;
            prev=fs;
            fs=front;
        }
        
    }
    ListNode* reverseKGroup(ListNode* head, int k) {
        if(head==nullptr || head->next==nullptr) return head;
        ListNode* temp=head;
        ListNode* fs=head;
        ListNode* prev_tail=nullptr;
        int ct=0;
        bool flag=true;
        while(temp){
            ++ct;
            if(ct==k){
                if(flag){
                    flag=false;
                    head=temp;
                }
                ListNode* nxt=temp->next;
                if(prev_tail) prev_tail->next=temp;
                prev_tail=fs;
                reverse(fs,nxt,k);
                fs=nxt;
                temp=nxt;
                ct=0;
            }
            else temp=temp->next;
        }
        return head;
    }
};
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