模拟

方法一递归

参考代码

  1. # Definition for singly-linked list.
  2. # class ListNode:
  3. #     def __init__(self, val=0, next=None):
  4. #         self.val = val
  5. #         self.next = next
  6. class Solution:
  7.     def mergeTwoLists(self, list1: Optional[ListNode], list2: Optional[ListNode]) -> Optional[ListNode]:
  8.         if not list1:
  9.             return list2
  10.         if not list2:
  11.             return list1
  13.         merge = None
  14.         if list1.val < list2.val:
  15.             merge = ListNode(list1.val)
  16.             list1 = list1.next
  17.         else:
  18.             merge = ListNode(list2.val)
  19.             list2 = list2.next
  20.         merge.next = self.mergeTwoLists(list1,list2)
  21.         return merge

复杂度分析

时间复杂度 O(n + m) n和n是链表的长度
空间复杂度 O(n + m) 主要是递归调用栈的开销

方法二迭代

参考代码

  1. # Definition for singly-linked list.
  2. # class ListNode:
  3. #     def __init__(self, val=0, next=None):
  4. #         self.val = val
  5. #         self.next = next
  6. class Solution:
  7.     def mergeTwoLists(self, list1: Optional[ListNode], list2: Optional[ListNode]) -> Optional[ListNode]:
  8.         pre = ListNode()
  9.         cur = pre
  10.         while list1 or list2:
  11.             if not list1:
  12.                 cur.next = list2
  13.                 break
  14.             if not list2:
  15.                 cur.next = list1
  16.                 break
  17.             if list1.val < list2.val:
  18.                 cur.next = list1
  19.                 list1 = list1.next
  20.             else:
  21.                 cur.next = list2
  22.                 list2 = list2.next
  23.             cur = cur.next
  24.         return pre.next

复杂度分析

时间复杂度 O(n + m) n和n是链表的长度
空间复杂度 O(1)