最有帮助的视频讲解
【LeetCode004-两个有序数组的中位数-最优算法代码讲解】 https://www.bilibili.com/video/BV1H5411c7oC/?share_source=copy_web&vd_source=afbacdc02063c57e7a2ef256a4db9d2a
时间复杂度$O(log(min(m,n)))$
空间复杂度是$S(1)$

class Solution {
public:
    double findMedianSortedArrays(vector<int>& nums1, vector<int>& nums2) {

        int m = nums1.size() , n = nums2.size();
        if(m > n){
            return findMedianSortedArrays(nums2,nums1);
        }
        int iMin = 0 , iMax = m;
        while(iMin <= iMax){
            int i = iMin + iMax >> 1;
            int j = (m + n + 1) / 2 - i;//公式
            if(i != 0 && j != n && nums1[i - 1 ] > nums2[j]){
                iMax = i - 1;
            }
            else if(j != 0 && i != m && nums2[j - 1] > nums1[i]){
                iMin = i + 1;
            }
            else {
                int maxLf = 0;
                if(i == 0)
                    maxLf = nums2[j - 1];
                else if(j == 0)
                    maxLf = nums1[i - 1];
                else
                    maxLf = max(nums1[i - 1],nums2[j - 1]);
                if((m + n)%2)
                    return maxLf;
                int maxRg = 0;
                if(i == m)
                    maxRg = nums2[j];
                else if(j == n)
                    maxRg = nums1[i];
                else 
                    maxRg = min(nums1[i],nums2[j]);
                return (maxLf + maxRg) / 2.0;
            }
        }
        return 0.0;

    }
};