Question

643. Maximum Average Subarray I

中文文档

Description

You are given an integer array nums consisting of n elements, and an integer k.

Find a contiguous subarray whose length is equal to k that has the maximum average value and return _this value_. Any answer with a calculation error less than 10-5 will be accepted.

 

Example 1:

Input: nums = [1,12,-5,-6,50,3], k = 4
Output: 12.75000
Explanation: Maximum average is (12 - 5 - 6 + 50) / 4 = 51 / 4 = 12.75

Example 2:

Input: nums = [5], k = 1
Output: 5.00000

 

Constraints:

• n == nums.length
• 1 <= k <= n <= 105
• -104 <= nums[i] <= 104

Solutions

Solution 1

class Solution:
  def findMaxAverage(self, nums: List[int], k: int) -> float:
    s = sum(nums[:k])
    ans = s
    for i in range(k, len(nums)):
      s += nums[i] - nums[i - k]
      ans = max(ans, s)
    return ans / k

class Solution {
  public double findMaxAverage(int[] nums, int k) {
    int s = 0;
    for (int i = 0; i < k; ++i) {
      s += nums[i];
    }
    int ans = s;
    for (int i = k; i < nums.length; ++i) {
      s += (nums[i] - nums[i - k]);
      ans = Math.max(ans, s);
    }
    return ans * 1.0 / k;
  }
}

function findMaxAverage(nums: number[], k: number): number {
  let n = nums.length;
  let ans = 0;
  let sum = 0;
  // 前k
  for (let i = 0; i < k; i++) {
    sum += nums[i];
  }
  ans = sum;
  for (let i = k; i < n; i++) {
    sum += nums[i] - nums[i - k];
    ans = Math.max(ans, sum);
  }
  return ans / k;
}

impl Solution {
  pub fn find_max_average(nums: Vec<i32>, k: i32) -> f64 {
    let k = k as usize;
    let n = nums.len();
    let mut sum = nums.iter().take(k).sum::<i32>();
    let mut max = sum;
    for i in k..n {
      sum += nums[i] - nums[i - k];
      max = max.max(sum);
    }
    f64::from(max) / f64::from(k as i32)
  }
}

class Solution {
  /**
   * @param Integer[] $nums
   * @param Integer $k
   * @return Float
   */
  function findMaxAverage($nums, $k) {
    $sum = 0;
    for ($i = 0; $i < $k; $i++) {
      $sum += $nums[$i];
    }
    $max = $sum;
    for ($j = $k; $j < count($nums); $j++) {
      $sum = $sum - $nums[$j - $k] + $nums[$j];
      $max = max($max, $sum);
    }
    return $max / $k;
  }
}