leetcode 396. Rotate Function 旋转向量的计算 + 寻找规律

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，但是直接双层循环会超时，为了找规律，先把具体的数字抽象为A,B,C,D，那么我们可以得到：

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]，可以写出代码如下：

代码如下：

#include <iostream>
#include <vector>
#include <map>
#include <unordered_map>
#include <set>
#include <unordered_set>
#include <queue>
#include <stack>
#include <string>
#include <climits>
#include <algorithm>
#include <sstream>
#include <functional>
#include <bitset>
#include <numeric>
#include <cmath>
#include <regex>

using namespace std;

class Solution
{
public:
int maxRotateFunction(vector<int>& a)
{
int t = 0, sum = 0, n = a.size();
for (int i = 0; i < n; ++i)
{
sum += a[i];
t += i * a[i];
}

int res = t;
for (int i = 1; i < n; ++i)
{
t = t + sum - n * a[n - i];
res = max(res, t);
}
return res;
}

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;
}
};