Reverse Nodes in k-Group

Description

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?

Code

Time Complexity: $O(n)$, Space Complexity: $O(1)$

Recursive

/**
 * 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* reverseKGroup(ListNode* head, int k) {
        ListNode* node = head;
        for(int i = 0; i < k; i++) {
            if(!node) return head;
            node = node->next;
        }
 
        ListNode* new_head = reverse(head, node);
        head->next = reverseKGroup(node, k);
        return new_head;
    }
 
    ListNode* reverse(ListNode* first, ListNode* last) {
        ListNode* prev = first;
        while(first != last) {
            ListNode* tmp = first->next;
            first->next = prev;
            prev = first;
            first = tmp;
        }
        return prev;
    }
};

Iterative

/**
 * 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* reverse_list(ListNode* first) {
        ListNode* prev = first;
        while(first) {
            ListNode* tmp = first->next;
            first->next = prev;
            prev = first;
            first = tmp;
        }
        return prev;
    }
 
    ListNode* reverseKGroup(ListNode* head, int k) {
        ListNode* dummy = new ListNode();
        dummy->next = head;
 
        ListNode* prev = dummy, *end, *nextHead;
        while(head) {
            end = head;
            int i;
            for(i = 0; i < k - 1 && end && end->next; end = end->next, i++);
            nextHead = end->next;
            if(i == k - 1) {
                end->next = nullptr;
                prev->next = reverse_list(head);
                prev = head;
                head->next = nextHead;
            }
            head = nextHead;
        }
 
        return dummy->next;
    }
};

Source

• Reverse Nodes in k-Group - LeetCode