has a very concise and fast solution based on binary search. Spend some time reading it and make sure you understand it. It is very helpful. 

Since the C++ interface of this problem has been updated, I rewrite the code below.

1 class Solution { 2 public: 3     double findMedianSortedArrays(vector
     
      & nums1, vector
      
       & nums2) { 4         int m = nums1.size(), n = nums2.size(); 5         if (m > n) return findMedianSortedArrays(nums2, nums1); 6         int i, j, imin = 0, imax = m, half = (m + n + 1) / 2; 7         while (imin <= imax) { 8             i = (imin & imax) + ((imin ^ imax) >> 1); 9             j = half - i;10             if (i > 0 && j < n && nums1[i - 1] > nums2[j]) imax = i - 1;11             else if (j > 0 && i < m && nums2[j - 1] > nums1[i]) imin = i + 1;12             else break;13         }14         int num1;15         if (!i) num1 = nums2[j - 1];16         else if (!j) num1 = nums1[i - 1];17         else num1 = max(nums1[i - 1], nums2[j - 1]);18         if ((m + n) & 1) return num1;19         int num2;20         if (i == m) num2 = nums2[j];21         else if (j == n) num2 = nums1[i];22         else num2 = min(nums1[i], nums2[j]);23         return (num1 + num2) / 2.0;24     }25 };