[Leetcode] 396. Rotate Function 解题报告

题目：

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 = 25F(1) = (0 * 6) + (1 * 4) + (2 * 3) + (3 * 2) = 0 + 4 + 6 + 6 = 16F(2) = (0 * 2) + (1 * 6) + (2 * 4) + (3 * 3) = 0 + 6 + 8 + 9 = 23F(3) = (0 * 3) + (1 * 2) + (2 * 6) + (3 * 4) = 0 + 2 + 12 + 12 = 26So the maximum value of F(0), F(1), F(2), F(3) is F(3) = 26.

思路：

1、暴力法：就是每个F函数的值都试一遍，最后得到最大值，思路比较直观。这样代码的复杂度就是O(n^2)了，果然无法通过大数据测试。

2、递推法：在网上看到一个非常巧妙的递归方程：F(i) = F(i - 1) + sum(A) - length(A) * A[length(A) - i]。这样就可以由F(i-1)在O(1)的时间内递推得到F(i)，将时间复杂度降低到O(n)。可是这个递推方程是怎么得到的呢？自己还是没有理解。

代码：

1、暴力法：

class Solution {public:    int maxRotateFunction(vector<int>& A) {        if (A.size() == 0) {            return 0;        }        int ans = INT_MIN;        for (int i = 0; i < A.size(); ++i) {            vector<int> B = rotate(A, i);            ans = max(ans, rotateFunction(B));        }        return ans;    }private:    vector<int> rotate(vector<int>& A, int k) {        if (A.size() == 0) {            return {};        }        if (k == 0) {            return A;        }        k = k % A.size();        vector<int> B(A);        rotate(B, 0, B.size() - 1);        rotate(B, 0, k - 1);        rotate(B, k, B.size() - 1);        return B;    }    void rotate(vector<int> &B, int start, int end) {        while (start < end) {            swap(B[start++], B[end--]);        }    }    int rotateFunction(vector<int>& B) {        int ans = 0;        for (int i = 0; i < B.size(); ++i) {            ans += i * B[i];        }        return ans;    }};

2、递推法：

class Solution {public:    int maxRotateFunction(vector<int>& A) {        int length = A.size(), sum_A = 0, sum = 0, ans = INT_MIN;        for (int i = 0; i < length; ++i) {            sum_A += A[i];            sum += i * A[i];        }        ans = sum;        for (int i = 1; i < length; ++i) {            sum = sum + sum_A - length * A[length - i];            ans = max(ans, sum);        }        return ans;    }};