Question

396. 旋转函数

English Version

题目描述

给定一个长度为 n 的整数数组 nums 。

假设 arrk 是数组 nums 顺时针旋转 k 个位置后的数组，我们定义 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 _。

生成的测试用例让答案符合 32 位 整数。

 

示例 1:

输入: nums = [4,3,2,6]
输出: 26
解释:
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 。

示例 2:

输入: nums = [100]
输出: 0

 

提示:

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

解法

方法一

class Solution:
  def maxRotateFunction(self, nums: List[int]) -> int:
    f = sum(i * v for i, v in enumerate(nums))
    n, s = len(nums), sum(nums)
    ans = f
    for i in range(1, n):
      f = f + s - n * nums[n - i]
      ans = max(ans, f)
    return ans

class Solution {
  public int maxRotateFunction(int[] nums) {
    int f = 0;
    int s = 0;
    int n = nums.length;
    for (int i = 0; i < n; ++i) {
      f += i * nums[i];
      s += nums[i];
    }
    int ans = f;
    for (int i = 1; i < n; ++i) {
      f = f + s - n * nums[n - i];
      ans = Math.max(ans, f);
    }
    return ans;
  }
}

class Solution {
public:
  int maxRotateFunction(vector<int>& nums) {
    int f = 0, s = 0, n = nums.size();
    for (int i = 0; i < n; ++i) {
      f += i * nums[i];
      s += nums[i];
    }
    int ans = f;
    for (int i = 1; i < n; ++i) {
      f = f + s - n * nums[n - i];
      ans = max(ans, f);
    }
    return ans;
  }
};

func maxRotateFunction(nums []int) int {
  f, s, n := 0, 0, len(nums)
  for i, v := range nums {
    f += i * v
    s += v
  }
  ans := f
  for i := 1; i < n; i++ {
    f = f + s - n*nums[n-i]
    if ans < f {
      ans = f
    }
  }
  return ans
}

function maxRotateFunction(nums: number[]): number {
  const n = nums.length;
  const sum = nums.reduce((r, v) => r + v);
  let res = nums.reduce((r, v, i) => r + v * i, 0);
  let pre = res;
  for (let i = 1; i < n; i++) {
    pre = pre - (sum - nums[i - 1]) + nums[i - 1] * (n - 1);
    res = Math.max(res, pre);
  }
  return res;
}

impl Solution {
  pub fn max_rotate_function(nums: Vec<i32>) -> i32 {
    let n = nums.len();
    let sum: i32 = nums.iter().sum();
    let mut pre: i32 = nums
      .iter()
      .enumerate()
      .map(|(i, &v)| (i as i32) * v)
      .sum();
    (0..n)
      .map(|i| {
        let res = pre;
        pre = pre - (sum - nums[i]) + nums[i] * ((n - 1) as i32);
        res
      })
      .max()
      .unwrap_or(0)
  }
}