# Definition for singly-linked list.
# class ListNode:
# def __init__(self, val=0, next=None):
# self.val = val
# self.next = next
class Solution:
def mergeKLists(self, lists: List[Optional[ListNode]]) -> Optional[ListNode]:
def merge(root1,root2):
if not root1 and not root2:
return None
elif not root1:
return root2
elif not root2:
return root1
else:
dummy = ListNode(-1)
tail = dummy
cur1 = root1
cur2 = root2
while cur1 and cur2:
if cur1.val <= cur2.val:
tail.next = cur1
cur1 = cur1.next
tail = tail.next
tail.next = None
else:
tail.next = cur2
cur2 = cur2.next
tail = tail.next
tail.next = None
if cur1:
tail.next = cur1
if cur2:
tail.next = cur2
return dummy.next
res = None
for head in lists:
res = merge(res,head)
return res
标签:ListNode,val,Python,self,链表,升序,return,root1,root2
From: https://www.cnblogs.com/DCFV/p/18417208