[LeetCode] 148. Sort List @ python

最新推荐文章于 2021-03-27 16:41:29 发布

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本文介绍了一种在O(nlogn)时间复杂度下对链表进行排序的方法，使用了归并排序的思想，通过快慢指针找到链表中点，并设计了合并两个已排序链表的函数。

一．题目：
在时间复杂度O(n log n) 的情况下对链表排序．
二．解题思路：
很明显，这道题希望我们利用归并排序的方法对链表进行排序．所以首先我们需要找到链表的中点，可以用快慢指针；
寻找链表中点的代码如下：

        #快慢指针寻找链表中点
        slow = head; fast = head                        
        while fast.next and fast.next.next:
            slow = slow.next
            fast = fast.next.next
        head1 = head
        head2 = slow.next
        slow.next = None

然后，还需要设计一个合并两个排序链表的函数，代码如下：

    def merge(self, head1, head2):
        if head1 == None: return head2
        if head2 == None: return head1
        dummy = ListNode(0)                             #归并时，新建一个链表头结点
        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 head1 == None:
            p.next = head2
        if head2 == None:
            p.next = head1
        return dummy.next

最终实现的代码如下所示：

# 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
        """
        if head == None or head.next == None:
            return head
        #快慢指针技巧的运用，用来截断链表。head1和head2为截为两条链表的表头
        slow = head; fast = head                        
        while fast.next and fast.next.next:
            slow = slow.next
            fast = fast.next.next
        head1 = head
        head2 = slow.next
        slow.next = None                                
        
        head1 = self.sortList(head1)
        head2 = self.sortList(head2)
        head = self.merge(head1, head2)
        return head
    
    def merge(self, head1, head2):
        if head1 == None: return head2
        if head2 == None: return head1
        dummy = ListNode(0)                             #归并时，新建一个链表头结点
        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 head1 == None:
            p.next = head2
        if head2 == None:
            p.next = head1
        return dummy.next