148. 排序链表_CodingPark编程公园

排序链表

问题

给你链表的头结点 head ，请将其按 升序 排列并返回 排序后的链表 。

进阶：
你可以在 O(n log n) 时间复杂度和常数级空间复杂度下，对链表进行排序吗？

示例 1：

输入：head = [4,2,1,3]
输出：[1,2,3,4]

示例 2：

输入：head = [-1,5,3,4,0]
输出：[-1,0,3,4,5]

示例 3：

输入：head = []
输出：[]

链接：https://leetcode-cn.com/problems/sort-list

解答

完成 进阶 要求 😁
归并排序 + 链表

# 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
        
        '''
        分
        '''
        slow, fast = head, head.next
        while fast and fast.next:
            fast = fast.next.next
            slow = slow.next
        # 切割
        mid, slow.next = slow.next, None
        left = self.sortList(head)
        right = self.sortList(mid)
        '''
        合
        '''
        res = dot = ListNode(0)
        while left and right:
            if left.val>right.val:
                dot.next = right
                right = right.next
            else:
                dot.next = left
                left = left.next
            dot = dot.next
        if left:
            dot.next = left
        else:
            dot.next = right
        
        return res.next

一点点感悟

看下面这段代码

# 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
        
        '''
        分
        '''
        slow, fast = head, head.next
        while fast and fast.next:
            fast = fast.next.next
            slow = slow.next
        # 切割
        mid, slow.next = slow.next, None
        left = self.sortList(head)
        right = self.sortList(mid)
        '''
        合
        '''
        res = dot = ListNode(0)
        while left and right:
            if left.val>right.val:
                dot.next = ListNode(right.val)
                right = right.next
            else:
                dot.next = ListNode(left.val)
                left = left.next
            dot = dot.next
        if left:
            dot.next = left
        else:
            dot.next = right
        
        return res.next

可以看出 new 一个 ListNode 十分消耗时间 🙅