25. Reverse Nodes in k-Group.md

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

• Java

  • mine

    Runtime: 0 ms, faster than 100.00%, Memory Usage: 39.7 MB, less than 8.62% of Java online submissions

    //O(N)time O(1)space
    public ListNode reverseKGroup(ListNode head, int k) {
        ListNode t = head;
        int size = 0;
        //get the node count
        while (t != null) {
            t = t.next;
            size++;
        }
        if (size < k || k == 1) {
            return head;
        }
    
        ListNode res = new ListNode(0);
        t = res;
        //need reverse count
        int reverseCount = size / k;
        int index = 0;
        ListNode reverse = null;
        ListNode temp;
        while (index < reverseCount) {
            size = 0;
            while (size < k) {
                temp = head.next;
                head.next = reverse;
                reverse = head;
                head = temp;
                size++;
            }
            t.next = reverse;
            while (t.next != null) {
                t = t.next;
            }
            reverse = null;
            index++;
        }
        t.next = head;
        return res.next;
    }
    

  • the most votes

    Runtime: 0 ms, faster than 100.00%, Memory Usage: 39.7 MB, less than 6.90% of Java online submissions

    //O(N)time O(1)space
    public ListNode reverseKGroup(ListNode head, int k) {
        ListNode curr = head;
        int count = 0;
        while (curr != null && count != k) { // find the k+1 node
            curr = curr.next;
            count++;
        }
        if (count == k) { // if k+1 node is found
            curr = reverseKGroup(curr, k); // reverse list with k+1 node as head
            // head - head-pointer to direct part, 
            // curr - head-pointer to reversed part;
            while (count-- > 0) { // reverse current k-group: 
                ListNode tmp = head.next; // tmp - next head in direct part
                head.next = curr; // preappending "direct" head to the reversed list 
                curr = head; // move head of reversed part to a new node
                head = tmp; // move "direct" head to the next node in direct part
            }
            head = curr;
        }
        return head;
    }