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发布于 2024-06-17 01:04:01 字数 4190 浏览 0 评论 0 收藏 0

396. Rotate Function

中文文档

Description

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

Solutions

Solution 1

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)
  }
}

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