【Leetcode】396. Rotate Function

题目链接：

题目：

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 = 25F(1) = (0 * 6) + (1 * 4) + (2 * 3) + (3 * 2) = 0 + 4 + 6 + 6 = 16F(2) = (0 * 2) + (1 * 6) + (2 * 4) + (3 * 3) = 0 + 6 + 8 + 9 = 23F(3) = (0 * 3) + (1 * 2) + (2 * 6) + (3 * 4) = 0 + 2 + 12 + 12 = 26So the maximum value of F(0), F(1), F(2), F(3) is F(3) = 26.思路：

啊 没想到这次contest中 我遇到最难的题是这道。。。用F(k)=F(k-1)-(n-1)*end+(sum-end)  + 0*end = F(k-1)+sum-n*end

画了个图：

算法：

[java] view plain copy

public int maxRotateFunction(int[] A) {

int max=Integer.MIN_VALUE;

int sum = 0;

int pre = 0;

for(int i=0;i<A.length;i++){

pre +=A[i]*i;

sum+=A[i];

}

max = Math.max(pre, max);

int k=1;

while(k<A.length){

int res = pre+sum-A.length*A[A.length-k];

max = Math.max(max, res);

pre = res;

k++;

}

return max;

}