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main.py
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main.py
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"""
23. Merge k Sorted Lists
You are given an array of k linked-lists lists, each linked-list is sorted in ascending order.
Merge all the linked-lists into one sorted linked-list and return it.
"""
class Solution:
def mergeKLists(self, lists: List[ListNode]) -> ListNode:
# return self.useHeap(lists)
return self.divideAndConquer(lists)
def divideAndConquer(self, lists: List[ListNode]) -> ListNode:
if not lists:
return None
if len(lists) == 1:
return lists[0]
def merge(left, right) -> ListNode:
dummyHead = ListNode()
head = dummyHead
while left or right:
if left and right:
if left.val < right.val:
head.next = left
head = head.next
left = left.next
else:
head.next = right
head = head.next
right = right.next
continue
if left:
head.next = left
head = head.next
left = left.next
continue
if right:
head.next = right
head = head.next
right = right.next
continue
return dummyHead.next
mid = len(lists) // 2
left = self.divideAndConquer(lists[:mid])
right = self.divideAndConquer(lists[mid:])
return merge(left, right)
# Time complexity: O(max(klogn)
def useHeap(self, lists: List[ListNode]) -> ListNode:
if not lists:
return None
heap = []
for i, node in enumerate(lists):
if node:
heap.append((node.val, i, node))
heapq.heapify(heap)
dummyNode = ListNode()
head = dummyNode
while heap:
val, i, node = heap[0]
if not node.next:
heapq.heappop(heap)
else:
heapq.heapreplace(heap, (node.next.val, i, node.next))
head.next = node
head = head.next
return dummyNode.next
# Time complexity: O(nlogn)
# Space complexity: O(n)
def bruteForce(self, lists: List[ListNode]) -> ListNode:
if not lists:
return None
result = []
for node in lists:
head = node
while head:
result.append(head.val)
head = head.next
result.sort()
dummyNode = ListNode()
head = dummyNode
for item in result:
head.next = ListNode(item)
head = head.next
return dummyNode.next