[Power button - small daily practice] 4. Looking for a median of two ordered arrays (python)

4. Looking for a median of two ordered arrays

Topic links: https://leetcode-cn.com/problems/median-of-two-sorted-arrays/

Title Description

Given two ordered arrays of size n, and m nums1and nums2.

Please find both the median and orderly array, and requires time complexity of the algorithm is O(log(m + n)).

You can assume nums1and nums2will not be both empty.

Examples

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

Code

Note: The following is what I wrote, seems wrong, I forgot to forget how time complexity, the time should be in charge of the point of view is wrong

class Solution:
    def findMedianSortedArrays(self, nums1: List[int], nums2: List[int]) -> float:
        nums = nums1 + nums2
        n = len(nums)
        for i in range(n):
            temp = nums[i]
            for j in range(i,n):
                if temp > nums[j]:
                    nums[i] = nums[j]
                    nums[j] = temp
                    temp = nums[i]
            
        if n % 2 == 0:
            return (nums[n//2 - 1]+nums[n//2])/2
        return nums[(n-1)//2]

Note: The following is not my writing, Baidu refer others.

class Solution:
    def findMedianSortedArrays(self, nums1: List[int], nums2: List[int]) -> float:
        m, n = len(nums1), len(nums2)
        if m > n:
            nums1, nums2, m, n = nums2, nums1, n, m
        if n == 0:
            raise ValueError
        imin, imax, half_len = 0, m, (m + n + 1) / 2
        while imin <= imax:
            i = int((imin + imax) / 2)
            j = int(half_len - i)
            if i < m and nums2[j-1] > nums1[i]:
                imin = i + 1
            elif i > 0 and nums1[i-1] >nums2[j]:
                imax = i - 1
            else:
                if i == 0: max_of_left = nums2[j-1]
                elif j == 0: max_of_left = nums1[i-1]
                else: max_of_left = max(nums1[i-1], nums2[j-1])
                if (m + n) % 2 == 1:
                    return max_of_left
                if i == m: min_of_right = nums2[j]
                elif j == n: min_of_right = nums1[i]
                else: min_of_right = min(nums1[i], nums2[j])
                return (max_of_left + min_of_right) / 2.0


# 作者：jutraman
# 链接：https://leetcode-cn.com/problems/median-of-two-sorted-arrays/solution/pythonliang-ge-you-xu-shu-zu-de-zhong-wei-shu-by-j/
# 来源：力扣（LeetCode）
# c著作权归作者所有。商业转载请联系作者获得授权，非商业转载请注明出处。