Added alternative solution using merge sort technique

Author: Aditya Mukherjee

Leetcode/Median_of_Two_Sorted_Arrays_MergeSort.cpp

@@ -0,0 +1,73 @@
+/**
+ * Approach: 
+    let nums1 = [1, 3, 4, 7, 10, 12], nums2 = [2, 3, 6, 15]
+
+    In order to find the median, we need a single sorted array. So a naive approach is that, just merge the 2 sorted arrays and find the median of that array.
+    This will have a time Complexity of O(n1 + n2), Space Complexity of O(n1 + n2)
+*/
+
+# include <bits/stdc++.h>
+
+class Solution {
+    std::vector<int> merge(std::vector<int>& nums1, std::vector<int>& nums2){
+        int n1 = nums1.size();
+        int n2 = nums2.size();
+
+        int i=0, j=0;
+
+        std::vector<int> res;
+        while(i < n1 and j < n2){
+            if(nums1[i] < nums2[j]){
+                res.push_back(nums1[i]);
+                i++;
+            } else {
+                res.push_back(nums2[j]);
+                j++;
+            }
+        }
+
+        while(i < n1){
+            res.push_back(nums1[i]);
+            i++;
+        }
+
+        while(j < n2){
+            res.push_back(nums2[j]);
+            j++;
+        }
+
+        return res;
+    }
+public:
+    double findMedianSortedArrays(std::vector<int>& nums1, std::vector<int>& nums2) {
+        std::vector<int> sortedarr = merge(nums1, nums2);
+
+        if(sortedarr.size() % 2 == 0){
+            return (sortedarr[sortedarr.size()/2] + sortedarr[sortedarr.size()/2 - 1])/2.0;
+        } else {
+            return (double)sortedarr[sortedarr.size()/2];
+        }
+    }
+};
+
+int main(int argc, char const* argv[]) {
+    int n;
+
+    std::cin >> n;
+    std::vector<int> arr1(n, 0);
+    for (int i = 0;i < n;i++) {
+        std::cin >> arr1[i];
+    }
+
+    std::cin >> n;
+    std::vector<int> arr2(n, 0);
+    for (int i = 0;i < n;i++) {
+        std::cin >> arr2[i];
+    }
+
+    Solution sol = Solution();
+    double res = sol.findMedianSortedArrays(arr1, arr2);
+
+    std::cout << res << "\n";
+    return 0;
+}