[Swift]LeetCode396. 旋转函数 | Rotate Function

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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 kpositions 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.

---

给定一个长度为 n 的整数数组 A 。

假设 Bk 是数组 A 顺时针旋转 k 个位置后的数组，我们定义 A 的“旋转函数” F 为：

F(k) = 0 * Bk[0] + 1 * Bk[1] + ... + (n-1) * Bk[n-1]。

计算F(0), F(1), ..., F(n-1)中的最大值。

注意:
可以认为 n 的值小于 105。

示例:

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

所以 F(0), F(1), F(2), F(3) 中的最大值是 F(3) = 26 。

---

60ms

class Solution {
    //F(i) = F(i-1) + sum - n*A[n-i]
    func maxRotateFunction(_ A: [Int]) -> Int {
        if A.isEmpty {return 0}
        var t:Int = 0
        var sum:Int = 0
        var n:Int = A.count
        for i in 0..<n
        {
            sum += A[i]
            t += i * A[i]
        }
        var res:Int = t
        for i in 1..<n
        {
            t = t + sum - n * A[n - i]
            res = max(res, t)
        }
        return res
    }
}

---

64ms

class Solution {
    func maxRotateFunction(_ A: [Int]) -> Int {
        // 假设 Bk 是数组 A 顺时针旋转 k 个位置后的数组
        // F(k) = F(k-1) + Sum(Bk) - Bk.count * Bk.last
        
        guard A.count > 1 else {
            return 0
        }
        
        let sum = A.reduce(0, {$0 + $1})
        var F0 = 0
        for i in 0..<A.count {
            F0 += i * A[i]
        }
        
        var res = F0
        var mx = res
        let n = A.count
        for i in 1..<A.count {
            res = res + sum - n * A[n-i]
            mx = max(mx, res)
        }
        
        return mx
    }
}