题解 | #斐波那契数列# | Rust

struct Solution{

}

impl Solution {
    fn new() -> Self {
        Solution{}
    }

    /**
    * 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
    *
    * 
        * @param n int整型 
        * @return int整型
    */
    pub fn Fibonacci(&self, n: i32) -> i32 {
        if n <= 2 {
            return 1;
        }
        let (mut a, mut b) = (1, 1);
        let mut ans = 0;
        for i in 3..n+1{
            ans = a + b;
            a = b;
            b = ans;
        }
        return ans;
    }
}