leetcode-javascript-50\. Pow(x,

50. Pow(x, n)

es6

function Pow(x,n){
return x**n;
}

//注意floor 
var myPow = function(x, n) {
    if(n<0){
        x = 1/x
        n = -n
    }
    let ans = 1;
    while(n>0) {
        if((n %2) == 1) {
            ans = ans * x;
        }
        x = x*x
        n = Math.floor(n/2)
        // console.log(n)
    }
    return ans
    // console.log(n)
};

1.暴力法

/**
 * @param {number} x
 * @param {number} n
 * @return {number}
 */
var myPow = function(x, n) {
    let N = n
    if(N < 0) {
        x = 1/x;
        N = -N
    }
    let ans = 1;
    for(let i = 0; i< N; i++) 
        ans = ans * x;
    return ans;
};
1.00000
2147483647
当n十分大时，会超时

2.递归快速幂

class Solution {
public:
    double fastPow(double x, long long n) {
        if (n == 0) {
            return 1.0;
        }
        double half = fastPow(x, n / 2);
        if (n % 2 == 0) {
            return half * half
        } else {
            return half * half * x;
        }
    }
    //主函数
    double myPow(double x, int n) {
        long long N = n;
        if (N < 0) {
            x = 1 / x; //对 根号x = 1/x
            N = -N; 
        }
        return fastPow(x, N);
    }
};
//迭代法
class Solution {
public:
    double myPow(double x, int n) {
        long long N = n;
        if (N < 0) {
            x = 1 / x;
            N = -N;
        }
        double ans = 1;
        double current_product = x;
        for (long long i = N; i >0; i /= 2) {
            //结束条件？？ 
            if ((i % 2) == 1) {
                ans = ans * current_product;
            }
            current_product = current_product * current_product;
        }
        return ans;
    }
};