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

(1) Java

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
class Solution {
    public ListNode reverseKGroup(ListNode head, int k) {
        ListNode dummy =  new ListNode(0);
        dummy.next = head;
        ListNode prev = dummy;
        ListNode next = null;
        ListNode p1 = head;
        while (p1 != null) {
            for (int i = 1; i < k; i++) {
                p1 = p1.next;
                if (p1 == null) {
                    return dummy.next;
                }               
            }
            next = p1.next;
            p1 = prev.next;
            ListNode p2 = p1.next;
            while(p2 != null && p2 != next) {
                ListNode p3 = p2.next;
                p2.next = p1;
                p1 = p2;
                p2 = p3;
            }
            ListNode p0 = prev.next;
            p0.next = next;
            prev.next = p1;
            prev = p0;
            p1 = next;
        }
        return dummy.next;
    }
}

(2) Python

# Definition for singly-linked list.
# class ListNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class Solution(object):
    def reverseKGroup(self, head, k):
        """
        :type head: ListNode
        :type k: int
        :rtype: ListNode
        """
        if not head:
            return head
        node = head
        for _ in xrange(1, k):
            node = node.next
            if not node:
                return head
        next = self.reverseKGroup(node.next, k)
        for _ in xrange(0, k):
            p1 = head.next
            head.next, next, head = next, head, p1
        return next

(3) Scala

Recursive

/**
 * Definition for singly-linked list.
 * class ListNode(var _x: Int = 0) {
 *   var next: ListNode = null
 *   var x: Int = _x
 * }
 */
object Solution {
    def reverseKGroup(head: ListNode, k: Int): ListNode = {
        if (head == null) {
            return head
        }
        var node = head
        for (i <- 1 until k) {
            node = node.next
            if (node == null) {
                return head
            }
        }
        var prev = reverseKGroup(node.next, k)
        var hdr = head
        for (i <- 0 until k) {
            val next = hdr.next
            hdr.next = prev
            prev = hdr
            hdr = next
        }
        return prev
    }
}

Iterative

/**
 * Definition for singly-linked list.
 * class ListNode(var _x: Int = 0) {
 *   var next: ListNode = null
 *   var x: Int = _x
 * }
 */
object Solution {
    def reverseKGroup(head: ListNode, k: Int): ListNode = {
        val dummy: ListNode = ListNode(0)
        dummy.next = head
        var prev: ListNode = dummy
        var p0 = head
        while (true) {
            var next = p0
            for (i <- 1 until k) {
                if (next == null) {
                    return dummy.next
                }
                next = next.next
            }
            if (next == null) {
                return dummy.next
            }
            next = next.next;
            for (i <- 0 until k) {
                var p1 = p0.next
                p0.next = next
                next = p0
                p0 = p1
            }
            val p2 = prev.next
            prev.next = next
            prev = p2
        }
        return dummy.next
    }
}