Problem description:

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.

Solution:

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1st k-1 loop init:
-1 -> 1 -> 2 -> 3 -> 4 -> 5
 |    |    |    
pre  cur  nex  

-1 -> 2 -> 1 -> 3 -> 4 -> 5
 |         |    |   
pre       cur  nex  

-1 -> 3 -> 2 -> 1 -> 4 -> 5
 |              |    |    
pre            cur  nex  


2nd loop init:
-1 -> 3 -> 2 -> 1 -> 4 -> 5
                |    |    |    
               pre  cur  nex

Above is how it works inside one group iteration(for example, k=3)

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/**
 * 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) {
        if(!head || k == 1) return head;
        ListNode *dummy= new ListNode(-1); dummy->next= head;
        ListNode *cur= dummy, *nex, *pre= dummy;
        
        //1. get the length of linked list
        int count= 0;
        while(cur= cur->next)
            count++;
        
        //2. use the length to do swapping
        while(count >= k){
            cur= pre->next;
            nex= cur->next;
            for(int i= 1; i< k; i++){ //IMPORTANT: change in k groups, only needs to swap k-1 times
                cur->next= nex->next;
                nex->next= pre->next;
                pre->next= nex;
                nex= cur->next;
            }
            pre= cur;
            count -= k;
        }
        return dummy->next;
    }
};

time complexity: $O(n)$
space complexity: $O(1)$
reference:
https://goo.gl/uyufCT