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

ゝ一世哀愁。 2022-06-04 06:27 165阅读 0赞

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;
        }
    };