25. Reverse Nodes in k-Group

• Difficulty: Hard

• Topics: Linked List

• Similar Questions:

  • Swap Nodes in Pairs

Problem:

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.

Example:

Given this linked list: 1->2->3->4->5

For k = 2, you should return: 2->1->4->3->5

For k = 3, you should return: 3->2->1->4->5

Note:

• Only constant extra memory is allowed.
• You may not alter the values in the list's nodes, only nodes itself may be changed.

Solutions:

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode* reverseKGroup(ListNode* head, int k) {
        ListNode* dummyHead = new ListNode(0);
        ListNode* tail = dummyHead;
        ListNode* left = head;
        int count = 0;
        while (head) {
            ++count;
            ListNode* nextNode = head->next;
            if (k == count) {
                tail->next = reverse(left, k);
                tail = left;
                left = nextNode;
                count = 0;
            }
            head = nextNode;
        }
        tail->next = left;
        return dummyHead->next;
    }

    ListNode* reverse(ListNode* head, int k) {
        ListNode* dummyHead = new ListNode(0);
        while (k-- > 0) {
            ListNode* nextNode = head->next;
            head->next = dummyHead->next;
            dummyHead->next = head;
            head = nextNode;
        }
        return dummyHead->next;
    }
};