<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:
    // Helper function to reverse a list
    ListNode *reverse(ListNode *thead)
    {
        ListNode *prev = nullptr;
        ListNode *curr = thead;

        while (curr != nullptr) {
            ListNode *nxt = curr->next;
            curr->next = prev;
            prev = curr;
            curr = nxt;
        }
        return prev;
    }

    ListNode* reverseKGroup(ListNode* head, int k) {
        if (!head || k == 1) return head;  
        int count = 0;
        ListNode *tcurr = head;
        ListNode *ans = nullptr;
        ListNode *aaa = head;
        ListNode *aaaa = nullptr;
        while (tcurr != nullptr) 
        {
            count++;
            tcurr = tcurr->next;
        }

        tcurr = head;  
        int z = k;

        while (count >= k) {
            ListNode *groupEnd = tcurr;
            for (int i = 1; i < k; i++) {
                groupEnd = groupEnd->next;
            }
            ListNode *nextGroup = groupEnd->next;
            groupEnd->next = nullptr;  
            ListNode *revv = reverse(tcurr);

            if (ans == nullptr) {
                ans = revv;  
                aaaa = revv;  
            } else {
                aaa->next = revv;  
                while (aaa->next != nullptr) {
                    aaa = aaa->next;
                }
            }

            aaa->next = nextGroup;
            tcurr = nextGroup; 

            count -= k; 
        }

        return aaaa;
    }
};

16200