JavaScriptor
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diff --git a/‎algorithms/cpp/rotateFunction/RotateFunction.cpp
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@@ -8,6 +8,7 @@ LeetCode
 
 | # | Title | Solution | Difficulty |
 |---| ----- | -------- | ---------- |
+|396|[Rotate Function](https://leetcode.com/problems/rotate-function/) | [C++](./algorithms/cpp/rotateFunction/RotateFunction.cpp)|Easy|
 |395|[Longest Substring with At Least K Repeating Characters](https://leetcode.com/problems/longest-substring-with-at-least-k-repeating-characters/) | [C++](./algorithms/cpp/longestSubstringWithAtLeastKRepeatingCharacters/LongestSubstringWithAtLeastKRepeatingCharacters.cpp)|Medium|
 |394|[Decode String](https://leetcode.com/problems/decode-string/) | [C++](./algorithms/cpp/decodeString/DecodeString.cpp)|Medium|
 |393|[UTF-8 Validation](https://leetcode.com/problems/utf-8-validation/) | [C++](./algorithms/cpp/UTF8Validation/UTF8Validation.cpp)|Medium|
 
@@ -0,0 +1,75 @@
+// Source : https://leetcode.com/problems/rotate-function/
+// Author : Hao Chen
+// Date   : 2016-11-03
+
+/*************************************************************************************** 
+ *
+ * Given an array of integers A and let n to be its length.
+ * 
+ * Assume Bk to be an array obtained by rotating the array A k positions clock-wise, we 
+ * define a "rotation function" F on A as follow:
+ * 
+ * F(k) = 0 * Bk[0] + 1 * Bk[1] + ... + (n-1) * Bk[n-1].
+ * 
+ * Calculate the maximum value of F(0), F(1), ..., F(n-1). 
+ * 
+ * Note:
+ * n is guaranteed to be less than 105.
+ * 
+ * Example:
+ * 
+ * A = [4, 3, 2, 6]
+ * 
+ * F(0) = (0 * 4) + (1 * 3) + (2 * 2) + (3 * 6) = 0 + 3 + 4 + 18 = 25
+ * F(1) = (0 * 6) + (1 * 4) + (2 * 3) + (3 * 2) = 0 + 4 + 6 + 6 = 16
+ * F(2) = (0 * 2) + (1 * 6) + (2 * 4) + (3 * 3) = 0 + 6 + 8 + 9 = 23
+ * F(3) = (0 * 3) + (1 * 2) + (2 * 6) + (3 * 4) = 0 + 2 + 12 + 12 = 26
+ * 
+ * So the maximum value of F(0), F(1), F(2), F(3) is F(3) = 26.
+ ***************************************************************************************/
+
+
+// 
+// Asumming we have 4 numbers: a, b, c, d, then
+//   F(0) = 0a + 1b + 2c + 3d
+//   F(1) = 3a + 0b + 1c + 2d
+//   F(2) = 2a + 3b + 0c + 1d
+//   F(3) = 1a + 2b + 3c + 0d
+//
+// We can see how F(n) transfrom to F(n+1) 
+//   F(0) - F(1) = -3a + b + c + d  
+//   F(1) - F(2) = a + -3b + c + d
+//   F(2) - F(3) = a + b + -3c + d
+//   F(3) - F(0) = a + b + c + -3d
+//   
+// So, we can tansfrom to the following experssion: 
+// 
+//   F(1) = F(0) - (a+b+c+d) + 4a  
+//   F(2) = F[1] - (a+b+c+d) + 4b
+//   F(3) = F[2] - (a+b+c+d) + 4c
+// 
+// Then, we can see this fomular:
+//
+//   F(n) = F(n-1) - sum(array) + len(array) * array[n-1]
+//   
+class Solution {
+public:    
+    int maxRotateFunction(vector<int>& A) {
+        int sum = 0;
+        int F = 0;
+        for (int i=0; i < A.size(); i++) {
+            sum += A[i];
+            F += (i * A[i]);
+        }
+        int maxF = F;
+        int len = A.size();
+        //cout << "F(0) = " << maxF <<endl;
+        for (int i=1; i< len; i++) {
+            F = F - sum + len * A[i-1];
+            //cout << "F(" << i << ") = " << F << endl;
+            maxF = max(maxF, F);
+        }
+        
+        return maxF;
+    }
+};