[leetcode] 396. Rotate Function 解题报告

题目链接：https://leetcode.com/problems/rotate-function/

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.

思路：根据F[0]，F[1]，F[2]，F[3]的规律可知F[1] = F[0] + sumOf(A) - 4*A[3];　依次类推．

代码如下：

class Solution {
public:
int maxRotateFunction(vector<int>& A) {
int len = A.size(), sum1 = 0, sum2 = 0, ans = INT_MIN;
for(int i = 0; i < len; i++) sum1 += A[i], sum2 += i*A[i];
ans = sum2;
for(int i = 1; i < len; i++)
{
sum2 = sum2 + sum1 - len*A[len-i];
ans = max(sum2, ans);
}
return ans;
}
};