LeetCode 396 Rotate Function 思维

You are given an integer array nums of length n.

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

• \(F(k) = 0 * arr_k[0] + 1 * arr_k[1] + ... + (n - 1) * arr_k[n - 1]\).

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.

Solution

假设 \(nums = [a,b,c,d], s = a+b+c+d\)，对于 \(F(0):\)

\[F(0) = b+2c+3d \]

对于 \(F(1):\)

\[F(1) = 3a+c+2d \]

所以：

\[F(1)-F(0) = 3a-b-c-d = 4a-s \]

其他也是同理。因此我们只需要递推即可求解所有的 \(F()\)， 复杂度 \(O(n)\)

点击查看代码

class Solution {
private:
    int ans = 0;
public:
    int maxRotateFunction(vector<int>& nums) {
        int s = 0, n = nums.size();
        for(int i=0;i<n;i++)s += nums[i];
        for(int i=0;i<n;i++) ans += i*nums[i];
        int f0 = ans;
        for(int i=1;i<n;i++){
            int fi = f0+n*nums[i-1]-s;
            ans = max(ans,fi);
            f0 = fi;
        }
        return ans;
    }
};