题目
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.

答案

class Solution {
    public int list_len(ListNode head) {
        int len = 0;
        ListNode curr = head;
        while(curr != null) {
            len++;
            curr = curr.next;
        }
        return len;
    }
    public ListNode reverseKGroup(ListNode head, int k) {
        int len = list_len(head);
        int groups = len / k;
        if(groups == 0) return head;

        // Keep track of start, reverse k guys, start again, until no start is availiable
        // prev_group_start: Start of previous group(before reversal), or End of previous group(after reversal)
        ListNode prev_group_start = new ListNode(0);
        ListNode start = head, ret = null;

        while(groups-- > 0) {
            // Reverse k nodes starting at start
            int cnt = 0;
            ListNode curr = start, prev_node = null;
            while(cnt < k && curr != null) {
                // Store curr.next
                ListNode next = curr.next;

                // Reverse
                curr.next = prev_node;
                
                // for the last guy, set prev_group_start.next to point to it
                if(++cnt == k || next == null) {
                    prev_group_start.next = curr;
                    if(ret == null) ret = curr;

                    // Update previous group start
                    prev_group_start = start;
                    // Temporarily point to the start of next group(in case next group has less than k elements)
                    prev_group_start.next = next;
                }
                prev_node = curr;
                curr = next;
            }
            start = curr;
        }
        return ret;
    }
}