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。
1 #-*-coding:utf-8-*- 2 3 class Node(object): 4 ''' 5 定义链表的结点 6 ''' 7 def __init__(self,val): 8 self.val = val 9 self.next = None 10 11 class Solution(object): 12 def initlist(self,nums): 13 ''' 14 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 15 ''' 16 l = Node(0) 17 ll = l 18 for num in nums: 19 node = Node(num) 20 ll.next = node 21 ll = ll.next 22 return l 23 24 def findMedianSortedArrays(self, nums1, nums2): 25 """ 26 :type nums1: List[int] 27 :type nums2: List[int] 28 :rtype: float 29 """ 30 list1 = self.initlist(nums1) 31 list2 = self.initlist(nums2) 32 list3 = Node(0) 33 pa = list1.next 34 pb = list2.next 35 pc = list3 36 37 while(pa and pb): 38 if(pa.val<=pb.val): 39 pc.next = Node(pa.val) 40 pc = pc.next 41 pa = pa.next 42 else: 43 pc.next = Node(pb.val) 44 pc = pc.next 45 pb = pb.next 46 47 # 插入剩余段 48 if pa: 49 pc.next = pa 50 else: 51 pc.next = pb 52 53 # 将已排好序的链表list3转换为列表nums3 54 nums3 = [] 55 ll = list3.next 56 while(ll): 57 nums3.append(ll.val) 58 ll = ll.next 59 60 j = len(nums3)/2 61 if len(nums3)%2==0: 62 return (nums3[j]+nums3[j-1])/2.0 63 else: 64 return nums3[j]