leetcode -- Sort List

https://leetcode.com/problems/sort-list/

这里又复习了一遍merge list的操作。用归并排序就行

参考http://www.cnblogs.com/zuoyuan/p/3699508.html

class Solution(object):
    def mergeList(self, head1, head2):

        if not head1: return head2
        if not head2: return head1

        dummy = ListNode(0)
        p = dummy

        while head1 and head2:

            if head1.val < head2.val:
                p.next = head1
                head1 = head1.next
                p = p.next
            else:
                p.next = head2
                head2 = head2.next
                p = p.next
        if not head1:
            p.next = head2
        if not head2:
            p.next = head1

        return dummy.next


    def sortList(self, head):
        """ :type head: ListNode :rtype: ListNode """

        if not head or not head.next: return head
        slow, fast = head, head

        while fast.next and fast.next.next:
            #这里要注意fast只需要到倒数第二个node就行，反正就只是将linkedlist分割而已。不需要准确二分之一
            #如果这里是while fast,那么可能fast就是None，然后slow是最后一个node，这样就导致linkedlist没有分成两半，出现死循环
            slow = slow.next
            fast = fast.next.next

        head1, head2 = head, slow.next
        slow.next = None
        head1, head2 = self.sortList(head1), self.sortList(head2)
        return self.mergeList(head1, head2)