148.排序链表.py

# 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

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