LeetCode148. SortList

题目描述：

Sort a linked list in O(n log n) time using constant space complexity.

解题思路：

排序不难，难的是在题目给定的条件下进行排序。

首先想到快排序。无奈发现单链表中交换元素很麻烦，放弃这个思路。接着想到归并排序，单链表中划分两个区域，归并，都很方便。

自顶向下的归并排序，空间复杂度为log(n)（n为链表长度）。于是想到自底向上归并排序，由于是用循环写的，空间复杂度为O(C)，满足题目要求。

解题代码：

class Solution(object):
    def sortList(self, head):
        if not head or not head.next:
            return head
        cur, length = head, 0
        while cur:
            length +=1
            cur = cur.next
        dummy = ListNode(0)
        dummy.next = head
        step = 1
        while step < length:
            tail, cur = dummy, dummy.next
            while cur:
                left, cur = self.split(cur, step)
                right, cur = self.split(cur, step)
                tail = self.merge(left, right, tail)
            step *= 2
        return dummy.next

    def split(self, cur, step):
        node = cur
        while cur and step > 1:
            cur = cur.next
            step -= 1
        if cur:
            tmp = cur.next
            cur.next = None
            cur = tmp
        return node, cur

    def merge(self, left, right, tail):
        while left and right:
            if left.val < right.val:
                tail.next = left
                tail, left = tail.next, left.next
            else:
                tail.next = right
                tail, right = tail.next, right.next
        if left:
            tail.next = left
        if right:
            tail.next = right
        while tail.next:
            tail = tail.next
        return tail

小结：

除了快排序，归并排序也很实用。