数值的整数次方-快速冪

方法一:暴力法

• 注意输入的指数的正负

  public class Solution {
    public double Power(double base, int exponent) {
        if(exponent == 0)
            return 1;
        if(exponent < 0)
        {
            base = 1/base;
        }
        int len = exponent > 0?exponent:-exponent;
        double res = base;
        for(int i = 0; i < len-1; i++)
        {
            res *= base;
        }
        return res;
  }
  }

  方法二:快速冪

• 位运算的优先级低于==

• java运算符优先级

  public class Solution {
    public double Power(double base, int exponent) {
        int len = exponent >= 0 ? exponent : -exponent;
        double res = 1.0;
        while(len != 0)
        {
            if((len&1)!=0)  //用位运算判断一个数的奇偶性
            {
                res *= base;
  
            }
            base *= base;
            len = len>>>1;
        }
        return exponent >= 0 ? res : 1/res;
  }
  }