Rust 力扣 - 2090. 半径为 k 的子数组平均值
文章目录
- 题目描述
- 题解思路
- 题解代码
- 题目链接
题目描述
题解思路
半径为 k 的子数组平均值 等价于 子数组长度为2 * k + 1的总和 除于 2 * k + 1
我们遍历长度为2 * k + 1的窗口，我们只需要记录窗口内的平均值即可
题解代码
impl Solution {
pub fn get_averages(nums: Vec<i32>, k: i32) -> Vec<i32> {
let n = ((k << 1) + 1) as usize;
let mut ans = vec![-1; nums.len()];
if n > ans.len() {
return ans;
}
let mut sum = 0;
for i in 0..n {
sum += nums[i] as usize;
}
for i in k as usize..(nums.len() - k as usize) {
ans[i] = (sum / n) as i32;
if i != nums.len() - k as usize - 1 {
sum += (nums[i + k as usize + 1] - nums[i - k as usize]) as usize;
}
}
ans
}
}
题目链接
https://leetcode.cn/problems/k-radius-subarray-averages/description/