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题目要求
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.
Bk 代表对数组 A 在位置 k 上进行顺时针的旋转后生成的数组。F(k) = 0 * Bk[0] + 1 * Bk[1] + … + (n-1) * Bk[n-1]，要求返回获得的最大的 F(k) 的值。
暴力循环
按照题目的要求，执行两次循环即可以获得 F(k) 的所有值，只需要从中比较最大值即可。
public int maxRotateFunction(int[] A) {
if(A == null || A.length == 0) return 0;
int max = Integer.MIN_VALUE;
for(int i = 0 ; i < A.length ; i++) {
int value = 0;
for(int j = 0 ; i < A.length ; j++) {
value += j * A[(j+i)%A.length];
}
max = Math.max(value, max);
}
return max;
}
数学思路
F(k) = 0 * Bk[0] + 1 * Bk[1] + … + (n-1) * Bk[n-1]
F(k-1) = 0 * Bk-1[0] + 1 * Bk-1[1] + … + (n-1) * Bk-1[n-1]
F(k) = F(k-1) + sum – n*Bk[0]
k = 0 Bk[0] = A[0]
k = 1 Bk[0] = A[len-1]
k = 2 Bk[0] = A[len-2]
…
public int maxRotateFunction(int[] A) {
if(A == null || A.length == 0) return 0;
int F = 0;
int sum = 0;
for(int i = 0 ; i<A.length ; i++) {
sum += A[i];
F += i * A[i];
}
int max = F;
for(int i = 1 ; i<A.length ; i++) {
F += sum – A.length * A[A.length – i];
max = Math.max(F, max);
}
return max;
}
