题解 | #幂#

function _pow(number,power) {
    // 补全代码
    //1: methods1 Math.pow()方法
    return Math.pow(number, power); //method1
    
    // 2: methods2 遍历相乘
    let x = 1;
    for(let i = 1; i <= power; i++) {
        x *= number;
    }
    return x;
    
    //3: methods3 递归  x^0 == 1
    return (power > 0) ? number * _pom(number, power -1) : 1/(number * _pom(number, power.abs() -1));
}