README.md

2074. Reverse Nodes in Even Length Groups

You are given the head of a linked list.

The nodes in the linked list are sequentially assigned to non-empty groups whose lengths form the sequence of the natural numbers (1, 2, 3, 4, ...). The length of a group is the number of nodes assigned to it. In other words,

• The 1st node is assigned to the first group.
• The 2nd and the 3rd nodes are assigned to the second group.
• The 4th, 5th, and 6th nodes are assigned to the third group, and so on.

Note that the length of the last group may be less than or equal to 1 + the length of the second to last group.

Reverse the nodes in each group with an even length, and return the head of the modified linked list.

Example 1:

Input: head = [5,2,6,3,9,1,7,3,8,4]
Output: [5,6,2,3,9,1,4,8,3,7]
Explanation:
- The length of the first group is 1, which is odd, hence no reversal occurs.
- The length of the second group is 2, which is even, hence the nodes are reversed.
- The length of the third group is 3, which is odd, hence no reversal occurs.
- The length of the last group is 4, which is even, hence the nodes are reversed.

Example 2:

Input: head = [1,1,0,6]
Output: [1,0,1,6]
Explanation:
- The length of the first group is 1. No reversal occurs.
- The length of the second group is 2. The nodes are reversed.
- The length of the last group is 1. No reversal occurs.

Example 3:

Input: head = [1,1,0,6,5]
Output: [1,0,1,5,6]
Explanation:
- The length of the first group is 1. No reversal occurs.
- The length of the second group is 2. The nodes are reversed.
- The length of the last group is 2. The nodes are reversed.

Constraints:

• The number of nodes in the list is in the range [1, 105].
• 0 <= Node.val <= 105

Solutions (Python)

1. Solution

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution:
    def reverseEvenLengthGroups(self, head: Optional[ListNode]) -> Optional[ListNode]:
        curr = head
        vals = []

        while curr is not None:
            vals.append(curr.val)
            curr = curr.next

        i = 0
        size = 1

        while i < len(vals):
            j = min(i + size, len(vals))
            if (j - i) % 2 == 0:
                vals[i:j] = vals[i:j][::-1]
            i += size
            size += 1

        curr = head

        for val in vals:
            curr.val = val
            curr = curr.next

        return head