leetcode 396. Rotate Function 旋转向量的计算 + 最笨的方法其实挺不错的

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.

题意很简单，就是做一次两层遍历即可，我认为这是最直接最简单的方法，

代码如下：

#include <iostream>#include <vector>#include <map>#include <set>#include <queue>#include <stack>#include <string>#include <climits>#include <algorithm>#include <sstream>using namespace std;class Solution {public:    int maxRotateFunction(vector<int>& a)    {        if (a.size() <= 0)            return 0;        long long allsum = 0;        long long sum2 = 0;        for (int i = 0; i < a.size(); i++) {            allsum += a[i] * i;            sum2 += a[i];        }        int maxRes = allsum;        for (int i = 0; i < a.size(); i++)         {            allsum -= sum2;            allsum += a[i];            allsum += a[i] * (int)(a.size() - 1);            maxRes = max((int)allsum, maxRes);        }        return maxRes;    }    int maxRotateFunctionByLoop(vector<int>& a)    {        if (a.size() <= 0)            return 0;        int maxRes = numeric_limits<int>::min();        for (int offset = 0; offset < a.size(); offset++)        {            int sum = 0;            for (int i = 0; i < a.size(); i++)            {                int index = (i + offset) % a.size();                sum += i * a[index];            }            maxRes = max(maxRes, sum);        }        return maxRes;    }};