[Leetcode]Median of Two Sorted Arrays

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
给出两个已经排序好的数组，算出两个数组的中位数
首先我的想法是先合并两个数组，然后根据个数，得到中位数的下表求出中位数

class Solution(object):    def findMedianSortedArrays(self, nums1, nums2):        """        :type nums1: List[int]        :type nums2: List[int]        :rtype: float        """        median = 0        nums = nums1 + nums2        m = len(nums)        if(m%2==0):            median=(nums[m/2]+nums[m/2+1])/2.0        else:            median=nums[m//2+1]        return(median)

虽然结果是对的，但是题目要求时间复杂度控制在O(log(m+n)),显然以上做法时间复杂度为O(m+n)。
然后网上看了别人用二分的思想

class Solution:    def findKthSortedArrays(self, A, B, k):        if len(A) < len(B):            tmp = A            A = B            B = tmp        if len(B) == 0:            return A[k - 1]        if k == 1:            return min(A[0], B[0])        pb = min(k / 2, len(B))        pa = k - pb        if A[pa - 1] > B[pb - 1]:            return self.findKthSortedArrays(A, B[pb:], k - pb)        elif A[pa - 1] < B[pb - 1]:            return self.findKthSortedArrays(A[pa:], B, k - pa)        else:            return A[pa - 1]    # @return a float    def findMedianSortedArrays(self, A, B):        if (len(A) + len(B)) % 2 == 1:            return self.findKthSortedArrays(A, B, (len(A) + len(B)) / 2 + 1)        else:            return (self.findKthSortedArrays(A, B, (len(A) + len(B)) / 2) +                self.findKthSortedArrays(A, B, (len(A) + len(B)) / 2 + 1)) / 2.0