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题目：

给你两个链表 list1 和 list2 ，它们包含的元素分别为 n 个和 m 个。
请你将 list1 中下标从 a 到 b 的全部节点都删除，并将list2 接在被删除节点的位置。
下图中蓝色边和节点展示了操作后的结果：

请你返回结果链表的头指针。

示例 1：

输入：list1 = [0,1,2,3,4,5], a = 3, b = 4, list2 = [1000000,1000001,1000002]
输出：[0,1,2,1000000,1000001,1000002,5]
解释：我们删除 list1 中下标为 3 和 4 的两个节点，并将 list2 接在该位置。上图中蓝色的边和节点为答案链表。

示例 2：

输入：list1 = [0,1,2,3,4,5,6], a = 2, b = 5, list2 = [1000000,1000001,1000002,1000003,1000004]
输出：[0,1,1000000,1000001,1000002,1000003,1000004,6]
解释：上图中蓝色的边和节点为答案链表。

代码：

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution:def mergeInBetween(self, list1: ListNode, a: int, b: int, list2: ListNode) -> ListNode:i = list1res = ListNode()temp = resfor _ in range(a):temp.next = itemp = temp.nexti = i.nextwhile list2:temp.next = list2temp = temp.nextlist2 = list2.nextfor _ in range(a,b+1):i = i.nexttemp.next = ireturn res.next

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution:def mergeInBetween(self, list1: ListNode, a: int, b: int, list2: ListNode) -> ListNode:i = list1temp = 0while list1:if temp == a-1:start = list1elif temp == b+1:end = list1list1 = list1.nexttemp += 1start.next = list2while list2.next:list2 = list2.nextlist2.next = endreturn i