题解 | #数值的整数次方#

public class Solution {
    public double Power(double base, int exponent) {
        if(exponent<0){
            base = 1/base;
            exponent *= -1;
        }
        double ans = 1;
        while(exponent>0){
            if((exponent&1)==1)
                ans = ans * base ;
            base = base * base;
            exponent = exponent>>1;
        }
        return ans;
  }
}

知识点：快速幂

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