题解 | #无环单链表插值#

自己写的半天通不过，最后请chatgpt帮写的，手动笑哭。。。

# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None
#
# 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
#
# 
# @param A int整型一维数组 
# @param val int整型 
# @return ListNode类
#
class Solution:
    def insert(self , A: List[int], val: int) -> ListNode:
        # write code here
        pre = ListNode(0)
        cur = pre
        n = len(A)
        for i in range(n):
            cur.next = ListNode(A[i])
            cur = cur.next
        head = pre.next
        node = ListNode(val)
        if head.val >= val:
            node.next = head
            return node
        prev, curr = head, head.next
        while curr and curr.val < val:
            prev, curr = curr, curr.next
        prev.next = node
        node.next = curr
        return head