# [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& 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); } }; ```