【leetcode】148. Sort List

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原文链接：http://www.cnblogs.com/seyjs/p/10552852.html

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#数据结构与算法

博客围绕链表排序问题展开，题目要求在O(n log n)时间复杂度和常数空间复杂度下对链表排序。解题采用快排方法，引入五个指针遍历链表分组，将小于、大于和等于基准值的节点分别存入不同链表，再对小链表和大链表递归，最后连接各链表。

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题目如下：

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

解题思路：因为题目要求时间复杂度是O(nlogn)，所以快排应该是合适的方法。原理也和快速排序差不多，我的方法是引入五个指针，分别是small_head,small_tail,large_head,large_tail,equal_tail，首先对list进行遍历，以head.val作为基准，小于head.val的保存到small的list里面，用small_head,small_tail分别指向small list的头结点和尾节点；大于head.val的保存到large list里面，用large_head,large_tail分别指向 list的头结点和尾节点；等于head.val的存入equal list，复用现有的head结点，equal_tail指向equal的尾节点。分组完成后，再对small list和large list做递归，最后把small list的尾节点指向equal list的head，同时把equal list的尾节点指向large list的头结点即可。

代码如下：

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

class Solution(object):
    def quick(self,sublist_head):
        if sublist_head == None:
            return (None,None)
        v = sublist_head.val
        previous = sublist_head
        current = sublist_head.next
        little_head = None
        little_current = None
        large_head = None
        large_current = None
        equal_current = sublist_head
        while current != None:
            if current.val < v:
                if little_head == None:
                    little_current = current
                    little_head = current
                    current = current.next
                    previous.next = current
                    little_current.next = None
                else:
                    little_current.next = current
                    little_current = little_current.next
                    current = current.next
                    previous.next = current
                    little_current.next = None
            elif current.val == v:
                equal_current.next = current
                previous = current
                equal_current = equal_current.next
                current = current.next
            else:
                if large_head == None:
                    large_head = current
                    large_current = current
                    current = current.next
                    previous.next = current
                    large_current.next = None
                else:
                    large_current.next = current
                    large_current = large_current.next
                    current = current.next
                    previous.next = current
                    large_current.next = None

        small_head,small_tail  =  self.quick(little_head)
        big_head,big_tail  =  self.quick(large_head)
        if small_head != None:
            small_tail.next = sublist_head
        equal_current.next = big_head
        return (small_head if small_head != None else sublist_head,big_tail if big_tail != None else equal_current)



    def sortList(self, head):
        """
        :type head: ListNode
        :rtype: ListNode
        """
        return self.quick(head)[0]