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
        numList = []
        p1 = head
        while p1:
            numList.append(p1.val)
            p1 = p1.next
        numList.sort()
        # print(numList)
        p1 = head
        count = 0
        while p1:
            p1.val = numList[count]
            count += 1
            p1 = p1.next
        return head

1. 链表排序很慢
2. 获取所有元素，用现有的函数功能进行排序
3. 将排序后的数据依次填入链表中