4. Median of Two Sorted Arrays

https://leetcode.com/problems/median-of-two-sorted-arrays/description/

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)).

nums1 = [1, 3]
nums2 = [2]

The median is 2.0

nums1 = [1, 2]
nums2 = [3, 4]

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

The hardest thing of this problem is to consider every border case. A good thought to solve this problem is dividing the two array into two equal-length parts,  and let max_of_left <= min_of_right, then median = (max_of_left + min_of_right) / 2.

class Solution(object):
    def findMedianSortedArrays(self, nums1, nums2):
        """
        :type nums1: List[int]
        :type nums2: List[int]
        :rtype: float
        """
        len1 = len(nums1)
        len2 = len(nums2)
        if len1 > len2:
            return self.findMedianSortedArrays(nums2, nums1)

        mid = (len1 + len2 + 1) / 2
        left = 0
        right = len1
        while left <= right:
            index1 = (left + right) / 2
            index2 = mid - index1
            if index1 < len1 and nums1[index1] < nums2[index2 - 1]:
                left = index1 + 1
            elif index1 > 0 and nums1[index1 - 1] > nums2[index2]:
                right = index1 - 1
            else:
                if index1 == 0:
                    max_of_left = nums2[index2 - 1]
                elif index2 == 0:
                    max_of_left = nums1[index1 - 1]
                else:
                    max_of_left = max(nums1[index1 - 1], nums2[index2 - 1])

                if (len1 + len2) % 2 == 1:
                    return max_of_left

                if index1 == len1:
                    min_of_right = nums2[index2]
                elif index2 == len2:
                    min_of_right = nums1[index1]
                else:
                    min_of_right = min(nums1[index1], nums2[index2])

                return float(max_of_left + min_of_right) / 2