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.
Solution
Two passes.
First pass, get the length of the list.
Second pass, swap in groups.
Time complexity: O(n)
Space complexity: O(1)
C++
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// Author: Huahua class Solution { public: ListNode *reverseKGroup(ListNode *head, int k) { if (!head || k == 1) return head; ListNode dummy(0); dummy.next = head; int len = 1; while (head = head->next) len++; ListNode* pre = &dummy; for (int l = 0; l + k <= len; l += k) { ListNode* cur = pre->next; ListNode* nxt = cur->next; for (int i = 1; i < k; ++i) { cur->next = nxt->next; nxt->next = pre->next; pre->next = nxt; nxt = cur->next; } pre = cur; } return dummy.next; } }; |
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