[LeetCode]396. Rotate Function

Easy

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.

分析如下：
F(1)-F(0)=(1-0)*A[0]+(2-1)*A[1]+(3-2)*A[2]+(0-3)*A[3]
F(2)-F(1)=(2-1)*A[0]+(3-2)*A[1]+(0-3)*A[2]+(1-0)*A[3]
F(3)-F(2)=(3-2)*A[0]+(0-3)*A[1]+(1-0)*A[2]+(2-1)*A[3]

sum = A[0]+…+A[length-1]
F(n)-F(n-1)=-(length-1)*A[length-n]+sum-A[length-n]=-length*A[length-n]+sum

F(n)=F(n-1)-length*A[length-n]+sum
故：
5ms:

public int maxRotateFunction(int[] A) {        if(A.length==0) return 0;        int max = Integer.MIN_VALUE;        int sum = 0,init = 0;        for(int i=0;i<A.length;i++)            sum += A[i];        for(int j=0;j<A.length;j++)            init += j*A[j];        max = init;        for(int k=1;k<A.length;k++){            init = init + sum -(A.length)*A[A.length-k];            max = Math.max(init, max);        }        return max;    }

如果暴力计算会超时：

public int maxRotateFunction2(int[] A) {        if(A==null||A.length==0) return 0;        int max = Integer.MIN_VALUE;        for(int i=0;i<A.length;i++){            int sum = 0;            for(int j=0;j<A.length;j++){                sum += ((i+j)%A.length) * A[j];            }            if(sum>max)                max = sum;        }        return max;    }