題目描述：在 O(n log n) 時間複雜度和常數級空間複雜度下，對連結串列進行排序。

示例 1:

輸入: 4->2->1->3
輸出: 1->2->3->4

示例 2:

輸入: -1->5->3->4->0
輸出: -1->0->3->4->5

解法1。歸併排序的遞迴做法，先找到連結串列的中點，然後從中間斷開，然後對前半段和後半段進行歸併排序，斷開這個操作是用遞迴，對前半段遞迴到底，回溯到倒數第二層，再對後半段遞迴到底，再呼叫merge函式。

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)
        '''
        # 上面這段等價於：
        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。其實是在上述做法的基礎上把merge函式改寫成了用迴圈做，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