Group_002

4. Median of Two Sorted Arrays

平凡解法：

先将 nums1 和 nums2 归并成一个数组 merged_arr ，然后再对归并后的数组求中位数。代码如下：

class Solution(object):
    def findMedianSortedArrays(self, nums1, nums2):
        """
        :type nums1: List[int]
        :type nums2: List[int]
        :rtype: float
        """
        m, n = len(nums1), len(nums2)
        i, j = 0, 0
        merged_arr = []
        while i <= m-1 and j <= n-1:
            if nums1[i] < nums2[j]:
                merged_arr.append(nums1[i])
                i += 1
            else:
                merged_arr.append(nums2[j])
                j += 1
        if i <= m-1:
            merged_arr.extend(nums1[i:])
        if j <= n-1:
            merged_arr.extend(nums2[j:])
        print(f'merged_arr = {merged_arr}')

        # When the `while` loop finished, we can get a merged array.
        # Then we just need to find the median of the merged array.
        if (m + n) % 2 == 1: # odd
            return merged_arr[(m+n)//2]
        else: # even
            return (merged_arr[(m+n)//2 - 1] + merged_arr[(m+n)//2]) / 2 # Note: accurate division and truncating division, 
                                                                         # see https://blog.youkuaiyun.com/feixingfei/article/details/7081446 for detail.


if __name__ == '__main__':
    sol = Solution()
    nums1 = [1, 2]
    nums2 = [3, 4]
    res = sol.findMedianSortedArrays(nums1, nums2)
    print(f'res = {res}')

这里面要注意三点：

• Python extend 方法：https://www.runoob.com/python/att-list-extend.html
• 归并排序：https://blog.youkuaiyun.com/weixin_45595437/article/details/105838132
• 在 Python2 中的除法是截断除法，而在 Python3 中才是精确除法，详见：https://blog.youkuaiyun.com/feixingfei/article/details/7081446

这种做法能够被 AC，但是它的时间复杂度是 $O(m+n)$ 。（参见： 归并排序时间复杂度分析）

要想让时间复杂度达到 $\log$ 级别，只能通过二分查找去做。如果用这种方法去做的话，那我们第一步要确定是要找一个数还是要找两个数。注意到 nums1 和 nums2 都是已经排过序的，这就为二分查找创造了条件。至此，这个问题就变成了“在两个数组中的二分查找问题”。这里边的关键是怎么样去压缩我们要查找的范围。为此，这里边要用到四个指针，它们分别指向 nums1 和 nums2 的头部和尾部：

nums1[lo1], nums1[hi1]
nums2[lo2], nums2[hi2]

时间复杂度为 $O(\log (\min (m,n))$ 的解法：（强烈推荐）

• YouTube. https://www.youtube.com/watch?v=LPFhl65R7ww&ab_channel=TusharRoy-CodingMadeSimple

Python 版本代码如下：

class Solution:
    def findMedianSortedArrays(self, nums1: List[int], nums2: List[int]) -> float:
        # If len(nums1) > len(nums2), then switch them so that nums1 is smaller than nums2.
        # This operation is used to decrease the time complexity so that the final time complexity 
        # will definitely be O(log(min(m, n))). Because once we guarantee nums1 is the smallest length, 
        # we always do binary search in the shortest list.
        if len(nums1) > len(nums2):
            nums1, nums2 = nums2, nums1
        
        x, y = len(nums1), len(nums2)
        lo, hi = 0, x # Note: here `hi` cannot be `x-1`
        while lo <= hi:
            partitionX = (lo + hi) // 2
            partitionY = (x + y + 1) // 2 - partitionX # Note: partitionX + partitionY = (x + y + 1) // 2
            
            # if partitionX is 0 it means nothing is there on left side. Use -INF for maxLeftX
            # if partitionX is the length of input then there is nothing on right side. Use +INF for minRightX
            maxLeftX = float('-inf') if partitionX == 0 else nums1[partitionX - 1]
            minRightX = float('inf') if partitionX == x else nums1[partitionX]
            
            maxLeftY = float('-inf') if partitionY == 0 else nums2[partitionY - 1]
            minRightY = float('inf') if partitionY == y else nums2[partitionY]
            
            if maxLeftX <= minRightY and maxLeftY <= minRightX:
                if (x + y) % 2 == 0:
                    return (max(maxLeftX, maxLeftY) + min(minRightX, minRightY)) / 2
                else:
                    return max(maxLeftX, maxLeftY)
            elif maxLeftX > minRightY: # we are too far on right side for partitionX. Go on left side.
                hi = partitionX - 1
            else: # we are too far on left side for partitionX. Go on right side.
                lo = partitionX + 1