Leetcode 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.

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

Analysis:

We need to keep a count of the node. If the count % k == 0, then we reverse the node between this range.

/**

* Reverse a link list between pre and next exclusively

* an example:

* a linked list:

* 0->1->2->3->4->5->6

* |           |

* pre        next

* after call pre = reverse(pre, next)

*

* 0->3->2->1->4->5->6

*          |  |

*          pre next

*

* @return the reversed list’s last node, which is the precedence of parameter next

Time complexity: O(n)

Space complexity: O(1)

Code is below:

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
public class Solution {
    public ListNode reverseKGroup(ListNode head, int k) {
        // O(n)
        
        if (head == null || head.next == null || k < 2) {
            return head;
        }
        
        ListNode dummy = new ListNode(0);
        ListNode prev = dummy;
        dummy.next = head;
        
        int count = 0;
        while (head != null) {
            count++;
            if (count % k == 0) {
                prev = reverseList(prev, head.next);
                head = prev.next;
            } else {
                head = head.next;
            }
        }
        return dummy.next;
    }
    
    // reverse list node from prev to end - 1
    public ListNode reverseList(ListNode prev, ListNode end) {
        ListNode cur = prev.next;
        ListNode temp = cur.next;
        
        while (temp != end) {
            cur.next = temp.next;
            temp.next = prev.next;
            prev.next = temp;
            temp = cur.next;
        }
        
        return cur;
    }
}