148排序链表

class ListNode:
    def __init__(self, x):
        self.val = x
        self.next = None
a = ListNode(1)
b = ListNode(3)
c = ListNode(2)
d = ListNode(1)
e = ListNode(5)
a.next = b
b.next = c
c.next = d
d.next = e
# 这道题是用归并排序的方法
class Solution:
    def sortList(self, head: ListNode) -> ListNode:
        # 首先判断链表是否只有一个节点，或者链表为空
        if not head or not head.next:return head
        # 定义两个快慢指针，用来找出链表的中间节点，用于等下递归的出口和入口
        slow,fast = head,head.next
        # 遍历链表，找出中间那个节点
        while fast and fast.next:
            slow,fast = slow.next,fast.next.next
        # 然后将链表分为两边，进行递归归并排序
        slow.next,mid = None,slow.next
        left = self.sortList(head)
        right = self.sortList(mid)
        # 定义一个节点，用来当做重新排序后链表的头结点
        node = ListNode(0)
        node1 = node
        # 遍历然后用于归并排序后的合成
        while left and right:
            if left.val > right.val:
                node.next,right  = right,right.next
            else:
                node.next,left = left,left.next
            node = node.next
        node.next = left if left else right
        # 最后返回这个链表。
        return node1.next


A = Solution()
print(A.sortList(a))