javascript 递归求阶乘

var n = parseInt(window.prompt('input'));


    function mul(n) {
        if (n === 1 || n === 0) {
            return 1
        }
        return n * mul(n - 1)
    }

    console.log(mul(n));

n的阶乘 计算公式: n* n-1的阶乘;

输入5
mul(5) => 5* mul(4) => 5 * 24 = 120
mul(4) => 4* mul(3) => 4 * 6 = 24
mul(3) => 3* mul(2) => 3 * 2 = 6
mul(2) => 2* mul(1) => 2 * 1 = 2

已知1的阶乘是1 将这个条件作为出口