题解 | #单链表的排序#

# 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 head ListNode类 the head node
# @return ListNode类
#
class Solution:
    def sortInList(self , head: ListNode) -> ListNode:
        # write code here
        if not head:
            return head
        l=[]
        while head:
            l.append(head.val)
            head=head.next
        l.sort()
        res=link=ListNode(0)
        for i in l:
            link.next=ListNode(i)
            link=link.next
        return res.next