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.
这道题是LeetCode第四次比赛的第一道题，博主第一道题就没有做出来，博主写了个O(n2)的方法并不能通过OJ的大数据集合，后来网上看大家的解法都是很好的找到了规律，可以在O(n)时间内完成。现在想想找规律的能力真的挺重要，比如之前那道Elimination Game也靠找规律，而用傻方法肯定超时，然后博主发现自己脑子不够活，很难想到正确的方法，说出来全是泪啊T.T。好了，来解题吧，我们为了找规律，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：
sum = 1A + 1B + 1C + 1D
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：
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;
}
};
参考资料：
https://leetcode.com/problems/rotate-function/
https://leetcode.com/problems/rotate-function/discuss/87853/Java-O(n)-solution-with-explanation
LeetCode All in One 题目讲解汇总(持续更新中…)
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