396. Rotate Function

题目

You are given an integer array nums of length n.

Assume arrk to be an array obtained by rotating nums by k positions clock-wise. We define the rotation function F on nums as follow:

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

Return the maximum value of F(0), F(1), ..., F(n-1).

The test cases are generated so that the answer fits in a 32-bit integer.

Example 1:

Input: nums = [4,3,2,6]
Output: 26
Explanation:
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.

Example 2:

Input: nums = [100]
Output: 0

Constraints:

n == nums.length
1 <= n <= 105
-100 <= nums[i] <= 100

题目大意

给定一个长度为n的整数数组nums，设arrk是数组nums顺时针旋转k个位置后的数组。

定义nums的旋转函数F为：

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

返回F(0), F(1), ..., F(n-1)中的最大值。

解题思路

抽象化观察：

nums = [A0, A1, A2, A3]

sum  = A0 + A1 + A2+ A3
F(0) = 0*A0 +0*A0 + 1*A1 + 2*A2 + 3*A3
    
F(1) = 0*A3 + 1*A0 + 2*A1 + 3*A2 
     = F(0) + (A0 + A1 + A2) - 3*A3 
     = F(0) + (sum-A3) - 3*A3 
     = F(0) + sum - 4*A3

F(2) = 0*A2 + 1*A3 + 2*A0 + 3*A1 
     = F(1) + A3 + A0 + A1 - 3*A2 
     = F(1) + sum - 4*A2

F(3) = 0*A1 + 1*A2 + 2*A3 + 3*A0 
     = F(2) + A2 + A3 + A0 - 3*A1 
     = F(2) + sum - 4*A1
    
// 记sum为nums数组中所有元素和
// 可以猜测当0 ≤ i < n时存在公式:
F(i) = F(i-1) + sum - n * A(n-i)

数学归纳法证明迭代公式：

根据题目中给定的旋转函数公式可得已知条件：

F(0) = 0×nums[0] + 1×nums[1] + ... + (n−1)×nums[n−1]；
F(1) = 1×nums[0] + 2×nums[1] + ... + 0×nums[n-1]。

令数组nums中所有元素和为sum，用数学归纳法验证：当1 ≤ k < n时，F(k) = F(k-1) + sum - n×nums[n-k]成立。

归纳奠基：证明k=1时命题成立。

F(1) = 1×nums[0] + 2×nums[1] + ... + 0×nums[n-1]
     = F(0) + sum - n×nums[n-1]

归纳假设：假设F(k) = F(k-1) + sum - n×nums[n-k]成立。

归纳递推：由归纳假设推出F(k+1) = F(k) + sum - n×nums[n-(k+1)]成立，则假设的递推公式成立。

F(k+1) = (k+1)×nums[0] + k×nums[1] + ... + 0×nums[n-1]
       = F(k) + sum - n×nums[n-(k+1)]

因此可以得到递推公式：

当n = 0时，F(0) = 0×nums[0] + 1×nums[1] + ... + (n−1)×nums[n−1]
当1 ≤ k < n时，F(k) = F(k-1) + sum - n×nums[n-k]成立。

循环遍历0 ≤ k < n，计算出不同的F(k)并不断更新最大值，就能求出F(0), F(1), ..., F(n-1)中的最大值。

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