25

25. Reverse Nodes in k-Group (Hard)

Given a linked list, reverse the nodes of a linked list k at a time and return its 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]

Example 3:

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

Example 4:

Input: head = [1], k = 1
Output: [1]

 

Constraints:

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

 

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

Companies:
Amazon, Microsoft, Facebook, Apple, eBay, Adobe, Google, Walmart Labs

Related Topics:
Linked List, Recursion

Similar Questions:

• Swap Nodes in Pairs (Medium)
• Swapping Nodes in a Linked List (Medium)

Solution 1.

// OJ: https://leetcode.com/problems/reverse-nodes-in-k-group/
// Author: github.com/lzl124631x
// Time: O(N)
// Space: O(1)
class Solution {
public:
    ListNode* reverseKGroup(ListNode* head, int k) {
        ListNode h, *tail = &h;
        while (head) {
            auto prev = tail;
            int i = 0;
            for (auto p = head; i < k && p; ++i, p = p->next);
            if (i < k) {
                tail->next = head;
                break;
            }
            for (int i = 0; i < k && head; ++i) {
                auto node = head;
                head = head->next;
                node->next = prev->next;
                prev->next = node;
            }
            while (tail->next) tail = tail->next;
        }
        return h.next;
    }
};

Solution 2.

// OJ: https://leetcode.com/problems/reverse-nodes-in-k-group/
// Author: github.com/lzl124631x
// Time: O(N)
// Space: O(1)
class Solution {
private:
    bool splitAtK(ListNode *head, int k, ListNode *&nextHead) {
        while (--k && head) head = head->next;
        if (!head) return false;
        nextHead = head->next;
        head->next = NULL;
        return true;
    }
    pair<ListNode*, ListNode*> reverseList(ListNode *head) {
        ListNode dummy(0), *tail = NULL;
        while (head) {
            ListNode *p = head;
            head = head->next;
            if (!tail) tail = p;
            p->next = dummy.next;
            dummy.next = p;
        }
        return { dummy.next, tail };
    }
public:
    ListNode* reverseKGroup(ListNode* head, int k) {
        ListNode *nextHead, dummy(0), *tail = &dummy;
        while (splitAtK(head, k, nextHead)) {
            auto p = reverseList(head);
            tail->next = p.first;
            tail = p.second;
            head = nextHead;
        }
        tail->next = head;
        return dummy.next;
    }
};