面试篇：Java 实现 x 的 y 次方

    前两天面试的一道题，记录一下：

Guava 中的实现：

/**
   * Returns {@code b} to the {@code k}th power. Even if the result overflows, it will be equal to
   * {@code BigInteger.valueOf(b).pow(k).intValue()}. This implementation runs in {@code O(log k)}
   * time.
   *
   * 
Compare {@link #checkedPow}, which throws an {@link ArithmeticException} upon overflow.
   *
   * @throws IllegalArgumentException if {@code k < 0}
   */
  @GwtIncompatible // failing tests
  public static int pow(int b, int k) {
    checkNonNegative("exponent", k); // 保证 k 为非负数
    switch (b) {
      case 0:
        return (k == 0) ? 1 : 0;
      case 1:
        return 1;
      case (-1):
        return ((k & 1) == 0) ? 1 : -1;
      case 2:
        return (k < Integer.SIZE) ? (1 << k) : 0;
      case (-2):
        if (k < Integer.SIZE) {
          return ((k & 1) == 0) ? (1 << k) : -(1 << k);
        } else {
          return 0;
        }
      default:
        // continue below to handle the general case
    }
    for (int accum = 1; ; k >>= 1) {
      switch (k) {
        case 0:
          return accum;
        case 1:
          return b * accum;
        default:
          accum *= ((k & 1) == 0) ? 1 : b;
          b *= b;
      }
    }
  }

网友实现方案，参见： 点击打开链接

public class Power {

    public static double pow(double a, int b){
        //check the validity of a and b
        if(a == 0 && b == 0)
            return Integer.MIN_VALUE;

        if(a == 0)
            return 0;

        if(b == 0)
            return 1;

        if(b == 1)
            return a;

        boolean aMinus = a < 0? true: false;
        boolean bMinus = b < 0? true : false;

        int bAbs = Math.abs(b);
        double aAbs = Math.abs(a);

        //check if b is odd
        double tempAnswer;
        if( (b & 1) != 0){
            tempAnswer = pow(aAbs, bAbs - 1) * aAbs;
        }
        else{
            tempAnswer = pow(aAbs * aAbs, bAbs / 2);
        }

        if(bMinus)
            tempAnswer = 1.0 / tempAnswer;
        if(aMinus && (b&1)!= 0)
        tempAnswer *= -1;

        return tempAnswer;

    }

    public static void main(String[] args){


        System.out.println(Power.pow(0,0));
        System.out.println(Power.pow(3,0));
        System.out.println(Power.pow(3,-1));
        System.out.println(Power.pow(3,100));
        return ;
    }
}

关键还是算法