396. Rotate Function

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Problem

You are given an integer array nums of length n.

Assume arrk to be an array obtained by rotating nums by k positions clock-wise. We define the rotation function F on nums as follow:

F(k) = 0 * arrk[0] + 1 * arrk[1] + … + (n - 1) * arrk[n - 1]. Return the maximum value of F(0), F(1), …, F(n-1).

The test cases are generated so that the answer fits in a 32-bit integer.

Example 1:

Input: nums = [4,3,2,6]
Output: 26
Explanation:
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.

Example 2:

Input: nums = [100]
Output: 0

Code

A,B,C,D，那么我们可以得到：

Sum = A + B + C + D

F(0) = 0A + 1B + 2C +3D

F(1) = 0D + 1A + 2B +3C

F(2) = 0C + 1D + 2A +3B

F(3) = 0B + 1C + 2D +3A

那么，我们通过仔细观察，我们可以得出下面的规律：

F(1) = F(0) + sum - 4D

F(2) = F(1) + sum - 4C

F(3) = F(2) + sum - 4B

那么我们就找到规律了, F(i) = F(i-1) + sum - n * A[n-i]

class Solution {
    public int maxRotateFunction(int[] nums) {
        int sum = 0;
        int f = 0;
        int n = nums.length;

        for(int i = 0; i < n; i++){
            sum += nums[i];
            f += i * nums[i];
        }

        int res = f;

        for(int i = 1; i < n; i++){
            f = f + sum - n * nums[n - i];
            res = Math.max(res, f);
        }

        return res;
    }
}