斐波那契

function fib1(n) {
    if (n <= 1) return n;
    return fib1(n - 2) + fib(n - 1);
}
// 最优解
function fib2(n) {
    if (n <= 1) return n;
    let first = 0;
    let second = 1;
    for(let i = 1; i < n; i++){
        second += first;
        first = second - first;
    }
    return second;
}