396. 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.

My code:
public int maxRotateFunction(int[] A) {
int n =A.length, result0=0, result, sum=0;
if (n==0) return 0;
for(int i=0;i<n;i++){
result0+=i*A[i];
sum +=A[i];
}
result = result0;
int[] sums = new int[n+1];
for(int i=1;i<=n; i++){
sums[i]= sums[i-1]+A[n-i];
}
for(int i=1;i<n;i++){
result = Math.max(result, result0+sum*i-n*sums[i]);
}
return result;
}总结：累加和。注意要找规律。另外，一开始直接遍历index，这样太慢了，O(N^2)，超时。
public int maxRotateFunction(int[] A) {
int allSum = 0;
int len = A.length;
int F = 0;
for (int i = 0; i < len; i++) {
F += i * A[i];
allSum += A[i];
}
int max = F;
for (int i = len - 1; i >= 1; i--) {
F = F + allSum - len * A[i];
max = Math.max(F, max);
}
return max;
}总结：别人的算法，更简洁一些。