leetcode 4 to find the median of two ordered arrays

Optimal Solution O (log (min (m, n)))

/ * * 
Before thinking combined with an ordered array made O ((m + n + 1 ) / 2), now try to O (log (min (m, n))) 
The basic idea is: Find than by binary the median obtain a corresponding array of small (assuming the presence of a last bounds of routine) 
assume respectively n1, n2, must n1 <= n2, two assumptions may median finally found that m1, m2 number (or assume present) 
when then binary search nums1, the initial value left = 0, right = n1; m1 is there [0, n1], m2 has a [k-n1, n1] ( k-n1> = 0 necessarily holds) 
and n1 <= n2, so certainly m2 between [0, n2], and therefore does not need to traverse the small array of things to worry about cross-border, just at the end of treatment 0 and n1, n2 of special circumstances; 
** * / 

class Solution {
 public :
     Double findMedianSortedArrays (Vector < int > & nums1, Vector < int > & nums2) {
         // first detects whether the first array is small, large arrays after 
        const  int N1 = nums1.size ();
         const  int N2 =nums2.size();
        if(n1>n2) return findMedianSortedArrays(nums2,nums1);
        
        int m1,m2;
        int left=0,right=n1,k=(n1+n2+1)/2;
        //二分查找m1,m2（假设存在）
        while(left<right){
            m1=left+(right-left)/2;
            m2=k-m1;
            if(nums1[m1]<nums2[m2-1])
                left=m1+1;
            else
                right=m1;
        } 
        M1 = left; 
        M2 = - K- M1; 
        
        int c1, C2;
         // find the first number c1, if the total is an odd number, returns 
        C1 = max (M1 <= 0 INT_MIN: nums1 [Ml-? . 1 ] , M2 <= 0 ? INT_MIN: nums2 [M2- . 1 ]);
         IF ((N1 + N2) & . 1 == . 1 )
             return C1; 
    
        // find the second number c2, and returns (c1 + c2) / 2 
        min = C2 (M1> N1 = INT_MAX: nums1 [M1], M2> = N2?? INT_MAX: nums2 [M2]);
         return  Double (C1 + C2) * 0.5 ; 
    } 
};

 

Method two: Merge ordered array

O(log(m+n+1)/2)

class Solution {
 public :
     Double findMedianSortedArrays (Vector < int > & nums1, Vector < int > & nums2) {
         // O (n-m + +. 1) / 2 Space O (n-) solution of time, size of the obtained two num1 (), respectively, with two pointers to two arrays, each of the small rearward displacement 
        int N1 = nums1.size (); int N2 = nums2.size ();
         int ID1, ID2;
         int MID = ( + N2 + N1 . 1 ) / 2 ;
         int I = 0 , J = 0 ;
         int CUR = INT_MIN;
         the while ! (MID = 0 && I <N1 && J <n2){
            mid--;
            if(nums1[i]<nums2[j])
                id1=nums1[i++];
            else
                id1=nums2[j++];
            
        }
        while(mid!=0&&i<n1){
            mid--;
            id1=nums1[i++];
        }
        while(mid!=0&&j<n2){
            mid--;
            id1=nums2[j++];
        }
        if((n1+n2)%2==1) return id1;
        id2=min((j>=n2?INT_MAX:nums2[j]),(i>=n1?INT_MAX:nums1[i]));
        return double(id1+id2)*0.5;
    }
};