力扣148-排序链表

对于nlogn的时间复杂度要求，使用归并，与数组不同的是：

1）链表找中点，通过快慢指针，找到左右的头节点即可，切断链表操作.

2）建立新的伪节点，不断比较left和right的值，进行组合。

3）这里空间复杂度还是有栈的调用，不是常数级别，后续代码太复杂，以后有时间再看

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution:
    def sortList(self, head: ListNode) -> ListNode:
        if not head or not head.next :return head
        # 1/ 切分链表，奇数节点数在中间，偶数节点数偏左
        slow = head
        fast = head.next
        while 1:
            if not fast or not fast.next:
                break
            fast = fast.next.next
            slow = slow.next
        mid = slow 
        rightNode = mid.next
        slow.next = None
        left,right = self.sortList(head),self.sortList(rightNode)
        # 归并排序
        h = res = ListNode(0)
        while left and right:
            if left.val<right.val:
                h.next = left
                left = left.next
            else:
                h.next = right
                right = right.next
            h = h.next
        h.next = left if left else right
        return res.next