Leetcode-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.

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 nodes, only nodes itself may be changed.

Only constant memory is allowed.

For 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

Have you met this question in a real interview?

 

Analysis:

Scan the whole list and find out the length len, then calculate the number of reverse iterations, which is len/k. For each iteration, reverse the corresponding group of nodes.

 

Solution:

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) {
 *         val = x;
 *         next = null;
 *     }
 * }
 */
public class Solution {
    public ListNode reverseKGroup(ListNode head, int k) {
        if (head==null || head.next==null) return head;
        if (k==0 || k==1) return head;

        ListNode preHead = new ListNode(0);
        preHead.next = head;
        ListNode cur = head;
        int len = 1;
        while (cur.next!=null){
            cur = cur.next;
            len++;
        }

        int iterNum = len/k;
        ListNode start = preHead;
        cur = head;
        for (int i=0;i<iterNum;i++){
            for (int j=0;j<k-1;j++){
                ListNode temp = cur.next.next;
                cur.next.next = start.next;
                start.next = cur.next;
                cur.next = temp;
            }

            start = cur;
            cur = cur.next;
        }

        return preHead.next;
    }
}

NOTE: Without counting the length of the list first,  we can also count whether there are k number of nodes left in the rest of list at the begnning of every reverse iteration.