【python/M/148】Sort List

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题目

基本思路

归并排序（应用快慢指针将链表一分为二）

实现代码

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

class Solution(object):
    def sortList(self, head):
        """
        :type head: ListNode
        :rtype: ListNode
        """
        # 归并排序
        if head == None or head.next == None:
            return head
        fast = slow = head
        while fast.next and fast.next.next:
            slow = slow.next
            fast = fast.next.next
        head1,head2 = head,slow.next
        slow.next = None
        head1 = self.sortList(head1)
        head2 = self.sortList(head2)
        head = self.mergeList(head1,head2)

        return head
    def mergeList(self,head1,head2):
        if head1 == None:
            return head2
        if head2 == None:
            return head1
        p = dummy = ListNode(0)

        while head1 and head2:
            if head1.val > head2.val:
                p.next = head2
                head2 = head2.next
                p = p.next
            else:
                p.next = head1
                head1 = head1.next
                p = p.next
        if head1 == None:
            p.next = head2
        if head2 == None:
            p.next = head1
        return dummy.next