CS/SICP in JS

연습문제 1.11

띵킹 2023. 2. 10. 16:50

재귀

function f(n){
    return n < 3 ? n 
            : f(n-1) + 2 * f(n-2) + 3 * f(n-3);
}

반복

function f_m(n) {
    return n < 3 ? n 
            : f_iter(2, 1, 0, n);
}

function f_iter(a, b, c, count){
    return count === 0
        ? a
        : f_iter(a + 2 * b + 3 * c, a, b, count -1)
}

 

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