将两个升序链表合并为一个新的 升序 链表并返回。新链表是通过拼接给定的两个链表的所有节点组成的。

示例 1：

输入：l1 = [1,2,4], l2 = [1,3,4]
输出：[1,1,2,3,4,4]

示例 2：

输入：l1 = [], l2 = []
输出：[]

示例 3：

输入：l1 = [], l2 = [0]
输出：[0]

提示：

  • 两个链表的节点数目范围是 [0, 50]
  • -100 <= node.val <="100
  • l1 和 l2 均按 非递减顺序 排列

非递归

增加一个节点

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# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution:
    def mergeTwoLists(self, list1: Optional[ListNode], list2: Optional[ListNode]) -> Optional[ListNode]:
        head = ListNode()
        p = head
        while list1 and list2:
            if list1.val <= list2.val:
                p.next = list1
                p = p.next
                list1 = list1.next
            else:
                p.next = list2
                p = p.next
                list2 = list2.next
        if not list1:
            p.next = list2
        else:
            p.next = list1
        return head.next

list2插入list1

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# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution:
    def mergeTwoLists(self, list1: Optional[ListNode], list2: Optional[ListNode]) -> Optional[ListNode]:
        ###第二种思路，在原有的list1基础上，将list2添加到list1
        """执行用时：44 ms, 在所有 Python3 提交中击败了43.43%的用户
        内存消耗：15.1 MB, 在所有 Python3 提交中击败了22.66%的用户"""
        p = ListNode(0,None)
        p.next = list1
        q = p
        while list1 is not None and list2 is not None:
            if list1.val > list2.val:
                q.next = ListNode(list2.val,list1)
                list2 = list2.next
                q = q.next
            else:
                q = q.next
                list1 = list1.next
        if list1 is None:
            q.next = list2
        return p.next

递归

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# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution:
    def mergeTwoLists(self, list1: Optional[ListNode], list2: Optional[ListNode]) -> Optional[ListNode]:
        ###第四个，递归
        """
        执行用时：32 ms, 在所有 Python3 提交中击败了96.98%的用户
        内存消耗：15.3 MB, 在所有 Python3 提交中击败了5.12%的用户
        """
        if list1 is None:
            return list2
        elif list2 is None:
            return list1
        else:
            if list1.val < list2.val:
                list1.next = self.mergeTwoLists(list1.next,list2)
                return list1
            else:
                list2.next = self.mergeTwoLists(list1,list2.next)
                return list2