Leetcode4:Median of Two Sorted Arrays@Python

There are two sorted arrays nums1 and nums2 of size m and n respectively.

Find the median of the two sorted arrays. The overall run time complexity should be O(log (m+n)).

Example 1:

nums1 = [1, 3]
nums2 = [2]

The median is 2.0

Example 2:

nums1 = [1, 2]
nums2 = [3, 4]

The median is (2 + 3)/2 = 2.5

这道题给定两个由小到大已排好序的列表，将它继续由从小到大的顺序排列，并返回新列表的中间值。（要求时间复杂度不能超过O(log (m+n))）

首先，给定的两个列表都是已经排好序的，所以我们不使用几种排序算法，他们的时间复杂度都超过了log(m+n)。我们可以使用数据结构中学到的归并(MergeList)两个已排好序的链表的方法。

 #-*-coding:utf-8-*-

 class Node(object):
     '''
     定义链表的结点
     '''
     def __init__(self,val):
         self.val = val
         self.next = None

 class Solution(object):
     def initlist(self,nums):
         '''
         链表转换函数，传入一个列表，返回列表中元素组成的有头结点的链表
         '''
         l = Node(0)
         ll = l
         for num in nums:
             node = Node(num)
             ll.next = node
             ll = ll.next
         return l

     def findMedianSortedArrays(self, nums1, nums2):
         """
         :type nums1: List[int]
         :type nums2: List[int]
         :rtype: float
         """
         list1 = self.initlist(nums1)
         list2 = self.initlist(nums2)
         list3 = Node(0)
         pa = list1.next
         pb = list2.next
         pc = list3

         while(pa and pb):
             if(pa.val<=pb.val):
                 pc.next = Node(pa.val)
                 pc = pc.next
                 pa = pa.next
             else:
                 pc.next = Node(pb.val)
                 pc = pc.next
                 pb = pb.next

         # 插入剩余段
         if pa:
             pc.next = pa
         else:
             pc.next = pb

         # 将已排好序的链表list3转换为列表nums3
         nums3 = []
         ll = list3.next
         while(ll):
             nums3.append(ll.val)
             ll = ll.next

         j = len(nums3)/2
         if len(nums3)%2==0:
             return (nums3[j]+nums3[j-1])/2.0
         else:
             return nums3[j]