算法:数值的整数次方

数值的整数次方

数值的整数次方

给定一个double类型的浮点数base和int类型的整数exponent。求base的exponent次方。

保证base和exponent不同时为0

分析

题目本质就是求快速幂

快速幂：

private double pow(double base, int exp) {
    if (exp == 0.0 || base == 1.0) return 1.0;
    double res = 1.0;
    while (exp != 0) {
        if ((exp & 1) == 1){
            res *= base;
        }
        base *= base;
        exp >>= 1;
    }
    return res;
}

代码

public class Solution {
    public double Power(double base, int exponent) {
        if (base == 0.0 && exponent < 0) return 0;

        boolean negative = exponent < 0.0;
        exponent = Math.abs(exponent);
        return negative ? 1 / helper(base, exponent) : helper(base, exponent);
    }

    private double helper(double base, int exp) {
        if (exp == 0.0 || base == 1.0) return 1.0;
        double res = 1.0;
        while (exp != 0) {
            if ((exp & 1) == 1){
                res *= base;
            }
            base *= base;
            exp >>= 1;
        }
        return res;
    }
}