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#
1
2
3
4
5
6
7
8
| 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
|
1
2
3
4
5
6
7
8
9
10
11
12
13
14
| 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;
}
}
|
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
| 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;
}
|
1
2
3
4
5
6
7
8
9
10
11
12
13
| 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)
}
}
|
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
| 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;
}
}
|