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

非正確答案，求的不是嚴格中位數，當兩個陣列元素個數總和為偶數時，求的是中間兩個數左邊的數：

class Solution:
    def findMedianSortedArrays(self, nums1, nums2):
        """
        :type nums1: List[int]
        :type nums2: List[int]
        :rtype: float
        """
        return self.findMedian_logn(nums1, len(nums1), nums2, len(nums2))

    def
 
 findMedian_logn(self, nums1, n1, nums2, n2):
        m1 = (n1-1)//2
        m2 = (n2-1)//2
        if n1==1:
            if n2 ==1:
                return nums1[0] if nums1[0]<nums2[0] else nums2[0]
            if n2%2==0:
                if nums1[0] >= nums2[m2+1]:
                    return nums2[m2+1
 
]
                elif nums1[0] <= nums2[m2]:
                    return nums2[m2]
                else:
                    return nums1[0]
            else:
                if nums1[0] >=nums2[m2]:
                    return nums2[m2]
                elif nums1[0] <= nums2[m2-1]:
                    return nums2[m2-1]
                else:
                    return nums1[0]
        elif n2==1:
            if n1%2==0:
                if nums2[0] >= nums1[m1+1]:
                    return nums1[m1+1]
                elif nums2[0] <= nums1[m1]:
                    return nums1[m1]
                else:
                    return nums2[0]
            else:
                if nums2[0] >= nums1[m1]:
                    return nums2[m2]
                elif nums2[0] <= nums1[m1-1]:
                    return nums1[m1-1]
                else:
                    return nums2[0]
        else:
            cutlen = n1/2 if n1/2 < n2/2 else n2/2
            if nums1[m1] == nums2[m2]:
                return nums1[m1]
            elif nums1[m1] < nums2[m2]:
                return self.findMedian_logn(nums1[cutlen:], n1-cutlen, nums2, n2-cutlen)
            else:
                return self.findMedian_logn(nums1, n1-cutlent, nums2[cutlen:], n2-cutlen)

leetcode接受的程式碼，比較暴力求解

class Solution:
    def findMedianSortedArrays(self, nums1, nums2):
        """
        :type nums1: List[int]
        :type nums2: List[int]
        :rtype: float
        """
        total = len(nums1)+len(nums2)
        res_list=[]  
        while len(nums1) and len(nums2):  
            if nums1[0]<nums2[0]:  
                res_list.append(nums1.pop(0))  
            else:  
                res_list.append(nums2.pop(0))  
        if len(nums1):  
            res_list+=nums1  
        elif len(nums2):  
            res_list+=nums2         

        if total%2==0:
            return (res_list[total//2-1]+res_list[total//2])/2
        else:
            return res_list[total//2]