# My code, non-recursive version, 30ms, with explanation O(n) complexity, O(1) space

• ``````class Solution {
public:
ListNode *reverseKGroup(ListNode *head, int k) {
ListNode dummy(0); // dummy head node to track the new head node
ListNode *dummyH = &dummy; // pointing to the last node of the k-node sublink (after reverse) just before the current sublink
ListNode *pre, *cur, *next, *start, *end;
int i;
do{
start = end = dummyH->next;// start points to the first node of the current k-node sublink, end points to the last one
for(i=0;i<k-1;i++)
{// move end to the last node of the current k-node sublink
end = end->next;
if(!end) return dummy.next; // if the current sublink has less than k nodes, just return
}
dummyH->next = end; // link the last node of the previous k-node sublink (after reverse) to the last node of the current sublink (before reverse)
dummyH =start; // update dummyH to starat, which is the last node of the current sublink after reverse
pre = start;
cur = pre->next;
while(pre!=end)
{ // reverse the current k-node sublink
next = cur->next;
cur->next = pre;
pre = cur;
cur = next;
}
dummyH->next = next; // link the last node (after reverse) to the next unprocessed node
}while(next);
return dummy.next;
}
};``````

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