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 or not head.next:
            return head
        
        def find_mid(head):
            slow = head
            fast = head.next
            while fast and fast.next:
                slow = slow.next
                fast = fast.next.next
            return slow

        def merge(l1:ListNode,l2:ListNode):
            dummy = ListNode(0)
            cur = dummy
            while l1 and l2:
                if l1.val<l2.val:
                    cur.next = l1
                    l1 = l1.next
                else:
                    cur.next = l2
                    l2 = l2.next
                cur = cur.next
            
            if l1:
                cur.next = l1
            if l2:
                cur.next = l2
            return dummy.next

        mid = find_mid(head)
        right_head = mid.next
        mid.next = None
        left = self.sortInList(head)
        right = self.sortInList(right_head)
        return merge(left,right)
适合使用归并排序