3392. Count Subarrays of Length Three With a Condition
https://leetcode.com/problems/count-subarrays-of-length-three-with-a-condition/
简单来说 任意三个数的子阵列切片
中间的数 要是两边相加除二
思考：
Sliding Windows
而且 Rust 居然有内建 .widnows() 函数
没内建的语言用
for i in 0..nums.len()-2 {
let a = nums[i];
let b = nums[i+1];
let c = nums[i+2];
// ...
}
Code:
impl Solution {
pub fn count_subarrays(nums: Vec<i32>) -> i32 {
nums.windows(3)
.filter(|w| w[1] % 2 == 0 && w[1] / 2 == w[0] + w[2])
.count() as i32
}
}