【剑指offer】面试题16 - 数值的整数次方

面试题16：数值的整数次方

题目描述：

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。

题目链接：LeetCode 牛客网

解法一：考虑周到拿offer

思路：

参数 base：

• 若是 0
• 若是 1

参数 exponent：

• 若是负数
• 若是 0
• 若是 1

class Solution {
    
    
public:
    double Power(double base, int exponent) {
    
    
        if(exponent == 0) return 1.0;
        if(exponent == 1) return base;
        if(base == 0.0) return 0.0;
        unsigned int absEx = (unsigned int)abs(exponent);
        double res = func( base, absEx);
        if(exponent < 0) res = (1.0/res);
        return res;
    }
    double func(double base, unsigned int exponent)
    {
    
    
        double res = 1.0;
        for(unsigned int i = 1; i <= exponent; i++)
            res *= base;
        return res;
    }
};

解法二：快速乘进行局部优化

思路：

求 a 的 n 次方公式：

class Solution {
    
    
public:
    double myPow(double base, int exponent) {
    
    
        if(exponent == 0) return 1.0;
        if(exponent == 1) return base;
        if(base == 0.0) return 0.0;
        if(base == 1.0) return 1.0;
        unsigned int absEx = (unsigned int)abs(exponent);
        double res = func( base, absEx);
        if(exponent < 0) res = (1.0/res);
        return res;
    }
    double func(double base, unsigned int e)
    {
    
    
        if(e == 0) return 1;
        if(e == 1) return base;
        double res = func(base, e >> 1);
        res *= res;
        if(e & 0x1 == 1) res *= base;
        return res;
    }
};