LeetCode 25. Reverse Nodes in k-Group

Linked List

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.

Follow up:

• Could you solve the problem in O(1) extra memory space?

• You may not alter the values in the list's nodes, only nodes itself 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

Solution:

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/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
    
    int length(ListNode* head) {
        int len = 0;
        while (head != nullptr) {
            head = head->next;
            len++;
        }
        return len;
    }
    
    ListNode* dfs(ListNode* head, int len, int k) {
        if (len < k) {
            return head;
        }
        ListNode* tail = head;
        
        ListNode* prev = nullptr, *tmp = nullptr;
        for (int i = 0; i < k; i++) {
            tmp = head->next;
            head -> next = prev;
            prev = head;
            head = tmp;
        }
        
        tail->next = dfs(head, len - k, k);
        return prev;
    }
    
public:
    ListNode* reverseKGroup(ListNode* head, int k) {
        return dfs(head, length(head), k);
    }
};