725. Split Linked List in Parts

Tags:

• Medium
• linked list
• practice

Link

It's not hard, but lots of edge cases

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

        dq = deque()
        current = head

        while current:
            dq.append(current)

            current = current.next

        reminder = len(dq) % k

        d = (len(dq) - reminder) // k

        res = [None for i in range(k)]

        for i in range(k):
            dummy = ListNode(-1)

            current = dummy

            for j in range(d):
                current.next = dq.popleft()
                current = current.next

            if i < reminder:
                current.next = dq.popleft()
                current = current.next

            current.next = None

            res[i] = dummy.next

        return res