[タイトル] LeetCodeブラシは、リストをソート

で  O（N-  ログイン  N-時の複雑さと一定の等級、リストの並べ替えのスペースの複雑さを）。

問題解決のアイデア：マージソート 

再帰：

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

class Solution(object):
    def sortList(self, head):
        """
        :type head: ListNode
        :rtype: 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

 

非再帰： 

# 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
        result = ListNode(0)
        result.next = head
        # 根据intv进行合并
        while intv < length:
            pre, h = result, result.next
            while h:
                # 得到待合并的两个list的头结点
                h1, i = h, intv
                while i and h:
                    h, i = h.next, i - 1
                if i:
                    break
                h2, i = h, intv
                while i and h:
                    h, i = h.next, i - 1
                c1, c2 = intv, intv - i
                # 合并
                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 result.next