25. Reverse Nodes in k-Group
1. Description
Given the head of a linked list, reverse the nodes of the list k at a time, and return the 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.
2. Example
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]
3. Constraints
- The number of nodes in the list is n.
- 1 <= k <= n <= 5000
- 0 <= Node.val <= 1000
4. Solutions
Linked List
n is the number of nodes in head
Time complexity: O(n)
Space complexity: O(1)
class Solution {
public:
ListNode *reverseKGroup(ListNode *head, int k) {
ListNode dummy(0, head);
ListNode *tail = &dummy;
while (head != nullptr) {
int count = 0;
for (auto iter = head; count < k && iter != nullptr; ++count) {
iter = iter->next;
}
if (count < k) {
break;
}
ListNode *prev = nullptr, *curr = head, *next = nullptr, *last = head;
for (int i = 0; i < k; ++i) {
next = curr->next;
curr->next = prev;
prev = curr;
curr = next;
}
tail->next = prev;
last->next = curr;
tail = last;
head = curr;
}
return dummy.next;
}
};