This link 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<int>& nums1, vector<int>& 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 };