Finding the median of two arrays by dichotomy

A daily topic ing, Today is a day hard topic 4. Median of Two Sorted Arrays

Method 1 ： Merge two arrays , Time complexity O(m+n), Spatial complexity O(m+n)

class Solution {
    
public:
    int findMin(vector<int>& nums) {
    
        int l=0,r=nums.size()-1,min=nums[0];
        while(l<=r){
    
            int mid=(l+r)/2;
            if(nums[mid]<min){
    
                min=nums[mid];
            }
            if(nums[l]==nums[mid]){
    
                l++;
                continue;
            }
            if(nums[l]<nums[mid]){
    
                if(nums[l]<min){
    
                    min=nums[l];
                }   
                l=mid+1;
            }
            else{
    
                if(nums[mid]<min){
    
                    min=nums[mid];
                }
                r=mid-1;
            }
            
        }
        return min;
    }
};

Method 2 ： Dichotomy , Time complexity O(log(m+n)), Spatial complexity O(1)

class Solution {
    
public:
    double findMedianSortedArrays(vector<int>& nums1, vector<int>& nums2) {
    
        int m=nums1.size(),n=nums2.size();
        int left=(m+n+1)/2,right=(m+n+2)/2;
        return (findK(nums1,0,nums2,0,left)+findK(nums1,0,nums2,0,right))/2.0;
    }

    int findK(vector<int>& nums1,int i,vector<int>& nums2,int j,int k){
    
        int m=nums1.size(),n=nums2.size();
        if(i>=m){
    
            return nums2[j+k-1];
        }
        if(j>=n){
    
            return nums1[i+k-1];
        }
        if(k==1){
    
            return min(nums1[i],nums2[j]);
        }
        int mid1=(i+k/2-1>=m)?INT_MAX:nums1[i+k/2-1];
        int mid2=(j+k/2-1>=n)?INT_MAX:nums2[j+k/2-1];
        if(mid1>mid2){
    
            return findK(nums1,i,nums2,j+k/2,k-k/2);
        }
        else{
    
            return findK(nums1,i+k/2,nums2,j,k-k/2);
        }
    }
};