LeetCode Reverse Nodes in k-Group

LeetCode Reverse Nodes in k-Group
LeetCode-solving-Reverse Nodes in k-Group original question

Flip the number of each k in a linked list without changing the number of k at the end.

Note:

Node values cannot be modified. Only the extra space of constants can be used.

Example:

Input: head = 1-> 2-> 3-> 4-> 5, k = 2
Output: 2-> 1-> 4-> 3-> 5

Input: head = 1-> 2-> 3-> 4-> 5, k = 3
Output: 3-> 2-> 1-> 4-> 5

Solutions

This is an upgraded version of Swap Nodes in Pairs. Let's take A look at the operations to flip k nodes, A-> B-> C-> D-> E. Now we need to flip BCD three nodes. Perform the following steps:

C-> BD-> CB-> EA-> D Return and Node B

We did two things above. We first flipped the k nodes (Steps 1 and 2) and connected them to the two nodes (Steps 3 and 4 ).

AC Source Code

# Definition for singly-linked list.class ListNode(object):    def __init__(self, x):        self.val = x        self.next = Noneclass Solution(object):    def reverseKGroup(self, head, k):        """        :type head: ListNode        :type k: int        :rtype: ListNode        """        if not head or k <= 1:            return head        dummy = ListNode(-1)        dummy.next = head        temp = dummy        while temp:            temp = self.reverseNextK(temp, k)        return dummy.next    def reverseNextK(self, head, k):        # Check if there are k nodes left        temp = head        for i in range(k):            if not temp.next:                return None            temp = temp.next        # The last node when the k nodes reversed        node = head.next        prev = head        curr = head.next        # Reverse k nodes        for i in range(k):            nextNode = curr.nex            curr.next = prev            prev = curr            curr = nextNode        # Connect with head and tail        node.next = curr        head.next = prev        return node

Please check out my Github for relevant source code.