题解 | 无环单链表插值

# 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
        A += [val]
        A.sort()
        head = ListNode(A[0])
        pre = head
        for x in A[1:]:
            cur = ListNode(x)
            pre.next = cur
            pre = cur
        return head
#先将数字合并到列表中后，再生成链表后返回