class Solution {
public:
double findMaxAverage(vector<int>& nums, int k) {
}
};
643. 子数组最大平均数 I
给你一个由 n 个元素组成的整数数组 nums 和一个整数 k 。
请你找出平均数最大且 长度为 k 的连续子数组，并输出该最大平均数。
任何误差小于 10-5 的答案都将被视为正确答案。
示例 1：
输入：nums = [1,12,-5,-6,50,3], k = 4 输出：12.75 解释：最大平均数 (12-5-6+50)/4 = 51/4 = 12.75
示例 2：
输入：nums = [5], k = 1 输出：5.00000
提示：
n == nums.length1 <= k <= n <= 105-104 <= nums[i] <= 104相似题目
原站题解
python3 解法, 执行用时: 208 ms, 内存消耗: 25.5 MB, 提交时间: 2023-09-27 15:27:27
class Solution:
def findMaxAverage(self, nums: List[int], k: int) -> float:
maxTotal = total = sum(nums[:k])
n = len(nums)
for i in range(k, n):
total = total - nums[i - k] + nums[i]
maxTotal = max(maxTotal, total)
return maxTotal / k
javascript 解法, 执行用时: 100 ms, 内存消耗: 52.5 MB, 提交时间: 2023-09-27 15:27:17
/**
* @param {number[]} nums
* @param {number} k
* @return {number}
*/
var findMaxAverage = function(nums, k) {
let sum = 0;
const n = nums.length;
for (let i = 0; i < k; i++) {
sum += nums[i];
}
let maxSum = sum;
for (let i = k; i < n; i++) {
sum = sum - nums[i - k] + nums[i];
maxSum = Math.max(maxSum, sum);
}
return maxSum / k;
};
cpp 解法, 执行用时: 140 ms, 内存消耗: 107.3 MB, 提交时间: 2023-09-27 15:26:59
class Solution {
public:
double findMaxAverage(vector<int>& nums, int k) {
int sum = 0;
int n = nums.size();
for (int i = 0; i < k; i++) {
sum += nums[i];
}
int maxSum = sum;
for (int i = k; i < n; i++) {
sum = sum - nums[i - k] + nums[i];
maxSum = max(maxSum, sum);
}
return static_cast<double>(maxSum) / k;
}
};
java 解法, 执行用时: 2 ms, 内存消耗: 53 MB, 提交时间: 2023-09-27 15:26:43
class Solution {
public double findMaxAverage(int[] nums, int k) {
int sum = 0;
int n = nums.length;
for (int i = 0; i < k; i++) {
sum += nums[i];
}
int maxSum = sum;
for (int i = k; i < n; i++) {
sum = sum - nums[i - k] + nums[i];
maxSum = Math.max(maxSum, sum);
}
return 1.0 * maxSum / k;
}
}
golang 解法, 执行用时: 204 ms, 内存消耗: 8.8 MB, 提交时间: 2021-07-01 10:41:02
func findMaxAverage(nums []int, k int) float64 {
ans, n := 0, len(nums)
for i := 0; i < k; i++ {
ans += nums[i]
}
s := ans
for i := 1; i <= n-k; i++ {
s += nums[i+k-1]-nums[i-1]
ans = max(ans, s)
}
return float64(ans)/float64(k)
}
func max(x, y int) int {
if x > y {
return x
}
return y
}