Leetcode: Reverse Nodes in k-Group

Reverse Nodes in k-Group

Similar Problems:

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.

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

Github: code.dennyzhang.com

Credits To: leetcode.com

Leave me comments, if you have better ways to solve.

## Blog link: https://code.dennyzhang.com/reverse-nodes-in-k-group
## Basic Ideas: Two pointer(p1, p2) with distance of k
##              Since head node might be changed, add a dummyNode
##              How to process:
##                 If p2 is None, stop changing
##                 Otherwise reverse list from p1 to p2
##              Move to next:
##                 p2 move to right with k distance
##                 If p2 is None, stop changing
## Complexity: Time O(n), Space O(1)
# 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 k <= 1:
            return head
        dummyNode = ListNode(None)
        dummyNode.next = head
        p1 = dummyNode

        while True:
            p2 = p1
            for i in xrange(k):
                if p2 is None:
                p2 = p2.next

            if p2 is None:

            # save the pointer of next p1
            q, s = p1.next, p2.next
            # reverse list from p1 to p2
            p = p1.next.next
            p1.next.next = s
            while p != s:
                r = p.next
                p.next = p1.next
                p1.next = p
                # move to next
                p = r
            p1 = q
        return dummyNode.next

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