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
/**
* Definition for singly-linked list.
* function ListNode(val) {
* this.val = val;
* this.next = null;
* }
*/
/**
* @param {ListNode} head
* @param {number} k
* @return {ListNode}
*/
var reverseKGroup = function(head, k) {
if (!head || k < 2) return head;
var count = 0;
var now = head;
var last = head;
var tmp = null;
while (now && count < k) {
now = now.next;
count++;
}
if (count === k) {
now = reverseKGroup(now, k);
while (count-- > 0) {
tmp = last.next;
last.next = now;
now = last;
last = tmp;
}
last = now;
}
return last;
};
Explain:
nope.
Complexity:
- Time complexity : O(n).
- Space complexity : O(1).