如何在不使用循环(如下所示)的情况下[91, 55, 77, 91]计算具有特定值(例如91)的向量(例如)中的元素?
fn count_eq(vec: &Vec<i64>, num: i64) -> i64 {
let mut counter = 0;
for i in vec {
if *i == num {
counter += 1;
}
}
return counter;
}
fn main() {
let v = vec![91, 55, 77, 91];
println!("count 91: {}", count_eq(&v, 91));
}
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lje*_*drz 11
你可以使用它Iterator::filter然后count:
fn main() {
let v = vec![91, 55, 77, 91];
println!("count 91: {}", v.iter().filter(|&n| *n == 91).count());
}
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