# 25. Reverse Nodes in k-Group
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
Approach 1: List manipulation.
# Code (Python)
Approach 1:
# Code (C++)
Approach 1:
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode(int x) : val(x), next(NULL) {}
* };
*/
// dummyHead->l1->l2->l3->l4->... (k = 2)
// | | | |
// left begin end fast
class Solution {
public:
ListNode* reverseKGroup(ListNode* head, int k) {
ListNode dummyHead(0);
ListNode *left = &dummyHead;
left->next = head;
ListNode *fast = head;
while (fast)
{
ListNode *begin = fast;
ListNode *end = left;
int i = 0;
for (; i < k && fast; ++i)
{
fast = fast->next;
end = end->next;
}
if (i < k)
break;
ListNode *prev = left;
ListNode *curr = begin;
while (curr != fast)
{
ListNode *next = curr->next;
curr->next = prev;
prev = curr;
curr = next;
}
left->next = end;
begin->next = fast;
left = begin;
}
return dummyHead.next;
}
};