[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