【LeetCode】148. 排序链表（字节算法一面）

一、题目描述：

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。

示例 1:

输入: 4->2->1->3
输出: 1->2->3->4

示例 2:

输入: -1->5->3->4->0
输出: -1->0->3->4->5

二、解题思路 & 代码

2.1 类似快排

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class Solution:
    def sortList(self, head: ListNode) -> ListNode:
        # 快速排序
        if not head: return None
        # 分成三个链表，分别是比轴心数小，相等，大的数组成的链表
        # small equal large 的缩写
        # 都指向相应链表的 head
        s = e = l = None
        target = head.val
        while head:
            nxt = head.next
            if head.val>target:
                head.next = l
                l = head
            elif head.val==target:
                head.next = e
                e = head
            else:
                head.next = s
                s = head
            head = nxt
        
        # 拆完各自排序即可，equal 无需排序
        s = self.sortList(s)
        l = self.sortList(l)
        # 合并 3 个链表， 这一步复杂度是 o(n)
        # # 可以同时返回链表的头指针和尾指针加速链表的合并。
        dummy = ListNode(0)
        cur = dummy # cur: 非 None 的尾节点
        # p: 下一个需要连接的节点
        for p in [s, e, l]:
            while p:
                cur.next = p
                p = p.next
                cur = cur.next
        return dummy.next

要注意的是，快速排序的最坏时间复杂度是 $o(n^{2})$ ，均摊复杂度才是 $o(nlogn)$。除非用 $bfprt$ 等算法先进行轴心数选择，才可以降低最坏时间复杂度。

参考：

1. LeetCode题解1

2.2 归并排序（递归法）

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class Solution:
    def sortList(self, head: ListNode) -> ListNode:
        if not head or not head.next: return head # termination.
        # cut the LinkedList at the mid index.
        slow, fast = head, head.next
        while fast and fast.next:
            fast, slow = fast.next.next, slow.next
        mid, slow.next = slow.next, None # save and cut.
        # recursive for cutting.
        left, right = self.sortList(head), self.sortList(mid)
        # merge `left` and `right` linked list and return it.
        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

2.3 归并排序（从底至顶直接合并）

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class Solution:
    def sortList(self, head: ListNode) -> ListNode:
        h, length, intv = head, 0, 1
        while h: h, length = h.next, length + 1
        res = ListNode(0)
        res.next = head
        # merge the list in different intv.
        while intv < length:
            pre, h = res, res.next
            while h:
                # get the two merge head `h1`, `h2`
                h1, i = h, intv
                while i and h: 
                    h, i = h.next, i - 1
                if i: 
                    break # no need to merge because the `h2` is None.
                h2, i = h, intv
                while i and h: 
                    h, i = h.next, i - 1
                c1, c2 = intv, intv - i # the `c2`: length of `h2` can be small than the `intv`.
                # merge the `h1` and `h2`.
                while c1 and c2:
                    if h1.val < h2.val: 
                        pre.next, h1, c1 = h1, h1.next, c1 - 1
                    else: 
                        pre.next, h2, c2 = h2, h2.next, c2 - 1
                    pre = pre.next
                pre.next = h1 if c1 else h2
                while c1 > 0 or c2 > 0: 
                    pre, c1, c2 = pre.next, c1 - 1, c2 - 1
                pre.next = h 
            intv *= 2
        return res.next

参考：

2. LeetCode题解2