add 396

README.md

@@ -246,6 +246,7 @@ My LeetCode solutions with Chinese explanation. 我的LeetCode中文题解。
 | 393 |[UTF-8 Validation](https://leetcode.com/problems/utf-8-validation/)|[C++](https://github.com/ShusenTang/LeetCode/blob/master/solutions/393.%20UTF-8%20Validation.md)|Medium|  |
 | 394 |[Decode String](https://leetcode.com/problems/decode-string/)|[C++](https://github.com/ShusenTang/LeetCode/blob/master/solutions/394.%20Decode%20String.md)|Medium|  |
 | 395 |[Longest Substring with At Least K Repeating Characters](https://leetcode.com/problems/longest-substring-with-at-least-k-repeating-characters/)|[C++](https://github.com/ShusenTang/LeetCode/blob/master/solutions/395.%20Longest%20Substring%20with%20At%20Least%20K%20Repeating%20Characters.md)|Medium|  |
+| 396 |[Rotate Function](https://leetcode.com/problems/rotate-function/)|[C++](solutions/396.%20Rotate%20Function.md)|Medium|  |
 | 397 |[Integer Replacement](https://leetcode.com/problems/integer-replacement/)|[C++](https://github.com/ShusenTang/LeetCode/blob/master/solutions/397.%20Integer%20Replacement.md)|Medium|  |
 | 398 |[Random Pick Index](https://leetcode.com/problems/random-pick-index/)|[C++](https://github.com/ShusenTang/LeetCode/blob/master/solutions/398.%20Random%20Pick%20Index.md)|Medium|  |
 | 400 |[Nth Digit](https://leetcode.com/problems/nth-digit)|[C++](https://github.com/ShusenTang/LeetCode/blob/master/solutions/400.%20Nth%20Digit.md)|Easy|  |

solutions/396. Rotate Function.md

@@ -0,0 +1,51 @@
+# [396. Rotate Function](https://leetcode.com/problems/rotate-function/)
+
+# 思路
+
+给定一个数组，然后给出F(i)的定义，求F(i)的最大值。
+此题最重要的就是根据定义快速求解出F(i)。
+
+先写出前几项（注意将A[i]对齐了）：
+``` 
+F(0) = 0*A[0] + 1*A[1] + 2*A[2] + 3*A[3] + ... + (n-2)*A[n-2] + (n-1) * A[n-1]
+F(1) = 1*A[0] + 2*A[2] + 3*A[3] + ... +          (n-1)*A[n-2] + 0*A[n-1]    
+F(2) = 2*A[0] + 3*A[2] + 4*A[3] + ... +              0*A[n-2] + 1*A[n-1]
+...
+```
+
+所以我们有
+```
+F(1) = F(0) + sum - A[n-1] - (n-1)*A[n-1];
+F(2) = F(1) + sum - A[n-2] - (n-1)*A[n-2];
+...
+
+F(i) = F(i-1) + sum - A[n-i] - (n-1)*A[n-i] = F(i-1) + sum - n * A[n-i];
+其中 sum = A[0] + A[1] + ... A[n-1]
+```
+
+所以我们可以先遍历一遍数组把sum和F(0)求出来, 再遍历一遍数组把每个F(i)都求出来同时保持一个全局最大值。
+
+注意可能会溢出，所以我们用long long型。
+
+两次遍历，时间复杂度O(n)，用滚动数组的思想可优化空间复杂度O(1)
+
+# C++
+``` C++
+class Solution {
+public:
+    int maxRotateFunction(vector<int>& A) {
+        long long sum = 0, f0 = 0, n = A.size();
+        for(int i = 0; i < n; i++){
+            sum += A[i];
+            f0 += i*A[i];
+        }
+        long long res = f0, fi = f0;
+        for(int i = 1; i < n; i++){
+            // fi = fi + sum - A[n-i] - (n-1)*A[n-i];
+            fi += (sum - n * A[n-i]);
+            res = max(res, fi);
+        }
+        return int(res);
+    }
+};
+```