1669. Merge two linked watches One question daily

topic：

1669. Merge two linked watches

Thought：

直接放上官方answer了，就yes模拟Linked。

topic要求将 list1
First a arrive b Delete all nodes，Replace it for list2
。therefore，我们首先找arrive list1
B a−1 Node preA，As well as b+1 Node aftB。because 1≤a≤b<n−1
（in n yes list1 length），so preA and aftB yes一定存在of。

Then we let us give up preA of next direction list2
of头节点，Give up list2
of尾节点of next direction aftB To。

Code：

class ListNode(object):
    def __init__(self, x):
        self.val = x
        self.next = None
class Solution:
    def mergeInBetween(self, list1: ListNode, a: int, b: int, list2: ListNode) -> ListNode:
        preA = list1
        for _ in range(a-1):
            preA = preA.next
        preB = preA
        for _ in range(b - a + 2):
            preB = preB.next
        preA.next = list2
        while list2.next:
            list2 = list2.next
        list2.next = preB
        return list1