4. 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 2.0Example 2:nums1 = [1, 2]nums2 = [3, 4]

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

Code

class Solution(object):    def findMedianSortedArrays(self, nums1, nums2):        """        :type nums1: List[int]        :type nums2: List[int]        :rtype: float        """        n1 = len(nums1)        n2 = len(nums2)        i = 0        j = 0        # sorting nums1 and nums2, sum pointer        sumpointer = 0        while (i < n1 or j < n2) and sumpointer <= (n1 + n2) >> 1:            if i >= n1:                j+=1            elif j >= n2:                i+=1            # combination nums1 and nums2 to sort            elif nums1[i] <= nums2[j]:                i+=1            else:                j+=1            sumpointer+=1        even = (n1 + n2) % 2 == 0        i-=1        j-=1        if i < 0:            return (nums2[j - 1] + nums2[j]) / 2.0 if even == True  else nums2[j]        if j < 0:            return (nums1[i - 1] + nums1[i]) / 2.0 if even == True else nums1[i]        # odd analysis        if even==False:            return max(nums1[i], nums2[j])        #even analysis， 1.shows stone point is in nums2        if nums1[i] < nums2[j]:            if j - 1 >= 0 and nums1[i] <= nums2[j - 1]:                return (nums2[j - 1] + nums2[j]) / 2.0            return (nums1[i] + nums2[j]) / 2.0        # shows stone point is in nums1        if i - 1 >= 0 and nums2[j] <= nums1[i - 1]:                return (nums1[i - 1] + nums1[i]) / 2.0        return (nums1[i] + nums2[j]) / 2.0