python - 运行时错误 - Leetcode 上的合并 k 个排序列表

Runtime Error TypeError: [] is not valid value for the expected return type ListNode raise TypeError(str(ret) + " is not valid value for the expected return type ListNode"); Line 65 in _driver (Solution.py) _driver() Line 71 in <module> (Solution.py)

# Definition for singly-linked list. # class ListNode(object): # def __init__(self, val=0, next=None): # self.val = val # self.next = next class Solution(object): def mergeKLists(self, lists): """ :type lists: List[ListNode] :rtype: ListNode """ if not lists: return lists if len(lists) == 1: return lists[0] # return itself if 'lists' is empty or only has one ListNode head = ListNode() dummy_front = head dummy_last = ListNode(10**5) # During comparison, if a linkedlist is out of nodes, assign dummy_last to be its last node. The maximum value of a listnode is 10**4, so dummy_last would be the largest.* flag = 0 # Use all_values to contain the current listnodes' values of all Linked lists. For example, with lists = [[1,4,5],[1,3,4],[2,6]], all_values = [1, 1, 2] all_values = [] for i in range(0, len(lists)): all_values.append(lists[i].val) while True: min_value = min(all_values) if min_value == 10**5: break # Finished merging of all linked lists min_index = all_values.index(min_value) if flag == 0: # To decide if it's the first node of our answer linked list to assign the "head" head.next = lists[min_index] dummy_front = lists[min_index] flag = 1 else: dummy_front.next = lists[min_index] dummy_front = dummy_front.next all_values.pop(min_index) # Pop the used smallest value if not lists[min_index].next: # If the linked list has no other nodes, assign dummy_last to be its last node. lists[min_index] = dummy_last else: lists[min_index] = lists[min_index].next all_values.insert(min_index, lists[min_index].val) # Insert new value at min_index return head.next

class Solution: def mergeKLists(self, lists: List[Optional[ListNode]]) -> Optional[ListNode]: vals = [] for node in lists: while node: vals.append(node.val) node = node.next if not vals: return vals.sort() result = ListNode(vals[0]) prev = result for v in vals[1:]: prev.next = ListNode(v) prev = prev.next return result

def mergeTwoLists(list1, list2): result = prehead = ListNode() while list1 and list2: if list1.val < list2.val: result.next = list1 list1 = list1.next else: result.next = list2 list2 = list2.next result = result.next result.next = list1 or list2 return prehead.next def mergeLists(lists): if len(lists) <= 2: # base case of recursion return lists[0] if len(lists) == 1 else mergeTwoLists(*lists) mid = len(lists) // 2 return mergeTwoLists(mergeLists(lists[:mid]), mergeLists(lists[mid:])) class Solution: def mergeKLists(self, lists): if lists: return mergeLists(lists)