leetcode oj java Rotate Function

一、问题描述：

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.

二、解决思路：

用一个数组记录每次rotate的id即可。

三、代码：

public class Solution {
public int maxRotateFunction(int[] A) {
if (A.length == 1 || A.length == 0) {
return 0;
}
if(A.length == 2){
return Math.max(A[0],A[1]);
}
int re = -2147483648;
int len = A.length;

int[] id = new int[len];
// init
for (int i = 0; i < len; i++) {
id[i] = i;
}

for (int i = 0; i < len; i++) {
int tmp = id[0];
id[0] = id[len - 1];
for (int t = len - 1; t > 1; t--) {
id[t] = id[t - 1];
}
id[1] = tmp;
tmp = 0;
for (int K = 0; K < len; K++) {
tmp += K * A[id[K]];
}
re = re > tmp ? re : tmp;
}

return re;
}
}