public class Solution { public double Power(double base, int exponent) { return exponent > 0 ? quickPow(base, exponent) : quickPow(1/base, -exponent); } public double quickPow(double base, int exp) { if(exp == 0) { return 1; } if(exp == 1) { return base; } return Power(base, exp/2)*Power(base, exp - exp/2); } }
因为所有数往下除以2最后都能被分解为1和0,所以对于0次幂,返回1,对于1次幂返回base,剩下的就除以二往下分解。
负数次幂就是把base变为1/base,然后-exp次幂即可