【题目描述】
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. 【解题思路】一开始想用暴力求解，也就是直接计算每次循环之后的结果找最大值，时间复杂度为O（n2），但是由于n比较大，所以直接TLE了
后来发现每次右移一个位置得到的新数组的sum值cursum是可以根据上一次的sum值prevsum计算得到的，公式为cursum=prevsum+allsum-sz*A[sz-i]，时间复杂度降为O(n)
【代码】
class Solution { public: int maxRotateFunction(vector<int>& A) { int sz=A.size(); if(sz<=1) return 0; //int sum[sz+1]={0}; int prevsum=0,maxsum=INT_MIN,allsum=0; for(int i=0;i<sz;i++){ prevsum+=A[i]*i; allsum+=A[i]; } maxsum=prevsum; for(int i=1;i<sz;i++){ int cursum=prevsum+allsum-sz*A[sz-i]; if(cursum>maxsum) maxsum=cursum; prevsum=cursum; } return maxsum; } };