Reverse Nodes in k-Group LeetCode Solution

Problem – Reverse Nodes in k-Group LeetCode Solution

Given the head of a linked list, reverse the nodes of the list k at a time, and return the modified list.

k 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 k then left-out nodes, in the end, should remain as it is.

You may not alter the values in the list’s nodes, only nodes themselves may be changed.

Example 1:

Input: head = [1,2,3,4,5], k = 2
Output: [2,1,4,3,5]

Example 2:

Input: head = [1,2,3,4,5], k = 3
Output: [3,2,1,4,5]

Constraints:

  • The number of nodes in the list is n.
  • 1 <= k <= n <= 5000
  • 0 <= Node.val <= 1000

Follow-up: Can you solve the problem in O(1) extra memory space?

Reverse Nodes in k-Group LeetCode Solution in Java

public ListNode reverseKGroup(ListNode head, int k) {
    ListNode curr = head;
    int count = 0;
    while (curr != null && count != k) { // find the k+1 node
        curr = curr.next;
        count++;
    }
    if (count == k) { // if k+1 node is found
        curr = reverseKGroup(curr, k); // reverse list with k+1 node as head
        // head - head-pointer to direct part, 
        // curr - head-pointer to reversed part;
        while (count-- > 0) { // reverse current k-group: 
            ListNode tmp = head.next; // tmp - next head in direct part
            head.next = curr; // preappending "direct" head to the reversed list 
            curr = head; // move head of reversed part to a new node
            head = tmp; // move "direct" head to the next node in direct part
        }
        head = curr;
    }
    return head;
}

Reverse Nodes in k-Group LeetCode Solution in C++

class Solution 
{
public:
    
    ListNode* reverse(ListNode* first, ListNode* last)
    {
        ListNode* prev = last;
        
        while ( first != last )
        {
            auto tmp = first->next;
            first->next = prev;
            prev = first;
            first = tmp;
        }
        
        return prev;
    }
    
    ListNode* reverseKGroup(ListNode* head, int k) 
    {
        auto node=head;
        for (int i=0; i < k; ++i)
        {
            if ( ! node  )
                return head; // nothing to do list too sort
            node = node->next;
        }

        auto new_head = reverse( head, node);
        head->next = reverseKGroup( node, k);
        return new_head;
    }
};

Reverse Nodes in k-Group LeetCode Solution in Python

def reverseKGroup(self, head, k):
    dummy = jump = ListNode(0)
    dummy.next = l = r = head
    
    while True:
        count = 0
        while r and count < k:   # use r to locate the range
            r = r.next
            count += 1
        if count == k:  # if size k satisfied, reverse the inner linked list
            pre, cur = r, l
            for _ in range(k):
                cur.next, cur, pre = pre, cur.next, cur  # standard reversing
            jump.next, jump, l = pre, l, r  # connect two k-groups
        else:
            return dummy.next
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