[LeetCode 396.] Rotate Function

LeetCode 396. Rotate Function

一道数学题。

题目描述

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.

解题思路

暴力做法 O(N^2) 是会超时的。注意观察给的例子，发现 F(k) 和 F(k-1) 之间存在相关关系，不用每次重新计算：
F(k) = F(k-1) + sum(A) - n * A[n - k]
由此，问题复杂度一下子降到了 O(N)。
发现递推关系，问题就简单了起来。

参考代码

要小心int类型溢出的问题。

/*
 * @lc app=leetcode id=396 lang=cpp
 *
 * [396] Rotate Function
 */

// @lc code=start
class Solution {
public:
/*
    int maxRotateFunction(vector<int>& A) {
        if (A.empty()) return 0;

        int res = INT32_MIN;
        int n = A.size();
        for (int k = 0; k < n; k++) {
            int F = 0;
            for (int i = 0; i < n; i++) {
                F += ((i + k) % n) * A[i];
            }
            res = max(res, F);
        }
        return res;
    } // TLE, brute force, O(N^2)
*/
    int maxRotateFunction(vector<int>& A) {
        if (A.empty()) return 0;

        size_t n = A.size();
        int64_t sum = 0; // avoid overflow
        for (int x : A) {
            sum += x;
        }
        int F0 = 0;
        for (int i = 0; i < n; i++) {
            F0 += i * A[i];
        }
        int res = F0;
        int64_t F = F0; // avoid overflow
        for (int k = 1; k < n; k++) {
            F = F + sum - n * A[n - k];
            res = max(res, (int)F);
        }
        return res;
    }
};
// @lc code=end