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
解法1。归并排序的递归做法，先找到链表的中点，然后从中间断开，然后对前半段和后半段进行归并排序，断开这个操作是用递归，对前半段递归到底，回溯到倒数第二层，再对后半段递归到底，再调用merge函数。
class Solution(object):
def sortList(self, head):
"""
:type head: ListNode
:rtype: ListNode
"""
if not head or not head.next:
return head
slow = head
pre = head
fast = head
while fast and fast.next:
pre = slow
slow = slow.next
fast = fast.next.next
pre.next = None
'''
l = sortList(head)
r = sortList(slow)
return self.merge(l, r)
'''
# 上面这段等价于：
return self.merge(self.sortList(head), self.sortList(slow))
def merge(self, l1, l2):
if not l1:
return l2
if not l2:
return l1
if l1.val < l2.val:
l1.next = self.merge(l1.next, l2)
return l1
else:
l2.next = self.merge(l1, l2.next)
return l2解法2。其实是在上述做法的基础上把merge函数改写成了用循环做，效率提高了不少
class Solution(object):
def sortList(self, head):
"""
:type head: ListNode
:rtype: ListNode
"""
if not head or not head.next:
return head
slow = head
pre = head
fast = head
while fast and fast.next:
pre = slow
slow = slow.next
fast = fast.next.next
pre.next = None
return self.merge(self.sortList(head), self.sortList(slow))
def merge(self, l1, l2):
if not l1:
return l2
if not l2:
return l1
dummy = ListNode(0)
cur = dummy
while l1 and l2:
if l1.val < l2.val:
cur.next = l1
l1 = l1.next
else:
cur.next = l2
l2 = l2.next
cur = cur.next
if l1:
cur.next = l1
if l2:
cur.next = l2
return dummy.next