(Leetcode) Median of Two Sorted Arrays (Hard)

 1 class Solution
 2 {
 3 public:
 4     double findMedianSortedArrays(vector<int>& nums1, vector<int>& nums2)
 5     {
 6         int m = nums1.size(), n = nums2.size();
 7         int res1 = findKth(nums1, nums2, (m + n + 1) / 2);
 8         int res2 = findKth(nums1, nums2, (m + n + 2) / 2);
 9         return (res1 + res2) / 2.0;
10     }
11 
12 private:
13     int findKth(vector<int> nums1, vector<int>nums2, int k)
14     {
15         int m = nums1.size(), n = nums2.size();
16         //always assume that m is equal or smaller than n
17         if (m > n) 
18             return findKth(nums2, nums1, k);
19         //nums1 equal null return nums2
20         if (m == 0) 
21             return nums2[k - 1];
22         if (k == 1) 
23             return min(nums1[0], nums2[0]);
24 
25         int i = min(m, k / 2), j = min(n, k / 2);
26         //divide k into two parts
27         if (nums1[i - 1] > nums2[j - 1])
28             return findKth(nums1, vector<int>(nums2.begin() + j, nums2.end()), k - j);
29         else if (nums1[i - 1] < nums2[j - 1])
30             return findKth(vector<int>(nums1.begin() + i, nums1.end()), nums2, k - i);
31         return 0;
32     }
33 };
34 
35 
36 int main()
37 {
38     Solution sol;
39     vector<int> a{ 1, 2, 3, 5 };
40     vector<int> b{ 6, 7, 8, 9 };
41     double res = sol.findMedianSortedArrays(a, b);
42     cout << res << endl;
43 
44     system("pause");
45     return 0;
46 }