leetcode148 Sort List

"""
Sort a linked list in O(n log n) time using constant space complexity.
Example 1:
Input: 4->2->1->3
Output: 1->2->3->4
Example 2:
Input: -1->5->3->4->0
Output: -1->0->3->4->5
"""
"""
这道题，主要分为两个步骤。
第一是用快慢指针法，找到链表的中间位置将链表一分为二
第二与leetcode21类似，https://www.cnblogs.com/yawenw/p/12273855.html
将两个链表合并为一个有序链表
整个过程采用递归的方法
"""

class ListNode:
    def __init__(self, x):
        self.val = x
        self.next = None

class Solution:
    def sortList(self, head):
        if not head or not head.next:
            return head
        cut, slow, fast = head, head, head  # cut指针用来截断链表
        while fast and fast.next:
            cut = slow
            slow = slow.next
            fast = fast.next.next
        cut.next = None  # bug 没写.next 理解截断的作用
        l1 = self.sortList(head)
        l2 = self.sortList(slow)
        return self.Merge(l1, l2)

    def Merge(self, l1, l2):
        head = ListNode(0)
        first = head  # 存结果的头指针
        while l1 and l2:
            if l1.val < l2.val:
                head.next = l1
                l1 = l1.next
            else:
                head.next = l2
                l2 = l2.next
            head = head.next  # !!!head随之向后移 容易忘
        head.next = l1 if l1 else l2
        # if not l2:
        #     move.next = l1
        # if not l1:
        #     move.next = l2
        return first.next