LeetCode 0396 - Rotate Function

Reck Zhang

发布于 2021-08-11 11:08:21

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发布于 2021-08-11 11:08:21

文章被收录于专栏：Reck ZhangReck Zhang

Rotate Function

Desicription

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.

Solution

class Solution {
public:
    long long maxRotateFunction(const std::vector<int>& A) {
        long long res = LONG_LONG_MIN;
        for(int i = 0; i < A.size(); i++) {
            long long sum = 0;
            for(int index = i, count = 0; count < A.size(); index = (index + 1) % A.size(), count++) {
                sum += A[index] * count;
            }
            res = std::max(res, sum);
        }
        return A.empty() ? 0 : res;
    }
};

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原始发表：2019-09-29，如有侵权请联系 cloudcommunity@tencent.com 删除

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