Sort List

LeetCode 148

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None


class Solution:
    def sortList(self, head):
        """
        :type head: ListNode
        :rtype: ListNode
        Time Complexity: O(n)
        """
        if not head or not head.next:
            return head

        slow = fast = head
        # 切對半 為何先 fast 在 fast.next 不能對調, 因為fast.next
        while fast and fast.next:
            fast = fast.next.next
            pre = slow
            slow = slow.next
        pre.next = None
        
        head2 = slow
        left = self.sortList(head)
        right = self.sortList(head2)
        return self.merge(left, right)

    def merge(self, left, right):
        # 對於 sinlge linked list 的 add 或 delete node 最好有 dummy node 會省事很多
        dummy = node = ListNode(0)

        while left and right:
            if left.val < right.val:
                node.next = left
                left = left.next
            else:
                node.next = right
                right = right.next

            node = node.next
        # if not left return right else return left
        node.next = left or right
        
        return dummy.next