LeetCode 396. Rotate Function

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).

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.

Note: n is guaranteed to be less than 105.

思考

如果不考虑优化，直接用两层循环来将每个F计算出来然后取最大值的话，复杂度为 O(n2) ,代码如下的第一个答案，会卡在第16个测试，超时，因此需要降低复杂度。

简化计算过程，使得计算一次F不需要一轮循环(n)。简化后的过程如下：
1. 计算F(0)和A的所有元素的和sum;
2. 通过F(0)计算F(A.size()-1)：F(0) - sum + A[0] * A.size()；
3. 与步骤2类似，通过F(A.size()-1)计算F(A.size()-2)，直到F(1)。

用这种方法的复杂度为 O(2n) 。

例子具体步骤如下表格：其中加粗的为 F(i), i = 0, A.size() - 1, A.size() - 2, ..., 1

A = [4, 3, 2, 6]

系数 4 3 2 6 和 last 0 1 2 3 25 -sum2 -1 0 1 2 10 + A[0] * A.size() 3 0 1 2 26 -sum2 2 -1 0 1 11 + A[1] * A.size() 2 3 0 1 23 …

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Q：为什么F(0)的下一个不是F(1)?
A：也可以按F(0),F(1)的顺序计算，只需要将- sum + A[i] * A.size()变成+ sum - A[i] * A.size()，并且将A从尾开始遍历即可。

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答案

c++
两层循环：

class Solution {public:    int maxRotateFunction(vector<int>& A) {        int length = A.size();        if (length == 0) return 0;        int max = -2147483648;        for (int i = 0; i < length; i++) {            int temp = 0;            for (int j = i, k = 0; k < length; k++, j = (j + 1) % length) {                temp += k * A[j];            }            if (temp > max) max = temp;        }        return max;    }};

一层循环：

class Solution {public:    int maxRotateFunction(vector<int>& A) {        int length = A.size();        if (length == 0) return 0;        int max = 0;  // result        int sum = 0;  // sum of A's elements        for (int i = 0; i < length; i++) {            max += i * A[i];  // F(0)            sum += A[i];        }        int last = max;  // last F        for (int i = 0; i < length - 1; i++) {            last -= sum;            last += A[i] * length;            max = last > max ? last : max;        }        return max;    }};