LeetCode 4. Find the median (hard) of two positive arrays

Given two sizes, they are m and n Positive order of （ From small to large ） Array nums1 and nums2. Please find and return the values of these two positive ordered arrays Median .

The time complexity of the algorithm should be O(log (m+n)) .

Example 1：

Input ：nums1 = [1,3], nums2 = [2]
Output ：2.00000
explain ： Merge array = [1,2,3] , Median 2
Example 2：

Input ：nums1 = [1,2], nums2 = [3,4]
Output ：2.50000
explain ： Merge array = [1,2,3,4] , Median (2 + 3) / 2 = 2.5

class Solution {
    
public:
    double findMedianSortedArrays(vector<int>& nums1, vector<int>& nums2) {
    
            // Use recursion to handle 
            int total = nums1.size() + nums2.size();
            if(total % 2 == 0)
            {
    
                int left = findKnum(nums1, 0, nums2, 0, total/2);
                int right = findKnum(nums1, 0, nums2,0, total/2 + 1);
                return (left + right) / 2.0;
            }
            else
                return findKnum(nums1, 0, nums2, 0, total/2 + 1);
    }

    int findKnum(vector<int>& nums1, int i, vector<int>& nums2, int j, int k)
    {
    
        // nums1  A long ,  Put the short one in front 
        if(nums1.size() - i > nums2.size() - j) 
            return findKnum(nums2, j, nums1, i, k);
        //  The left array is empty 
        if(nums1.size() == i) return nums2[j + k - 1];

        // k == 1  Directly return the minimum value of two arrays 
        if(k == 1) return min(nums1[i], nums2[j]);

        int si = min(i + k /2, int(nums1.size())), sj = j + k /2;
        if(nums1[si - 1] > nums2[sj - 1])
        {
    
            return findKnum(nums1, i, nums2, j+ k /2, k - k /2);
        }
        else
        {
    
            return findKnum(nums1, si, nums2,j, k - (si - i));
        }
    }
};