面试题16. 数值的整数次方(Leetcode) (pow函数实现)

面试题16. 数值的整数次方

难度中等4

实现函数double Power(double base, int exponent)，求base的exponent次方。不得使用库函数，同时不需要考虑大数问题。

 

示例 1:

输入: 2.00000, 10
输出: 1024.00000

示例 2:

输入: 2.10000, 3
输出: 9.26100

示例 3:

输入: 2.00000, -2
输出: 0.25000
解释: 2-2 = 1/22 = 1/4 = 0.25

class Solution {
public:
    double mp(double x , long n)
    {
        cout << n << ' ';
        if(n == 0)
            return 1;
        if(n == 1)
            return x;
        double save = mp(x,n/2);
        save *= save;
        if(n%2 == 1)
            return save*x;
        else
            return save;
    }
    double myPow(double x, int n) {
        double res = x;
        long num = n;
        if(n == 0)
            return 1;
        if(num < 0)
        {
            num = abs(num);
            res = 1/res;
        }
        cout << num << ' ' << res <