50. Pow(x, n)

问题：

实现x^n的幂运算。

Example 1:
Input: 2.00000, 10
Output: 1024.00000

Example 2:
Input: 2.10000, 3
Output: 9.26100

Example 3:
Input: 2.00000, -2
Output: 0.25000
Explanation: 2-2 = 1/22 = 1/4 = 0.25

Note:
-100.0 < x < 100.0
n is a 32-bit signed integer, within the range [−2^31, 2^31 − 1]

　　

解法：bit位方法

使用bit位方法，可以将计算速度提升一半

The basic idea is to decompose the exponent into powers of 2, so that you can keep dividing the problem in half. 

N = 9 = 2^3 + 2^0 = 1001 in binary. Then:
x^9 = x^(2^3) * x^(2^0)

从2^0->2^n 升序判断幂N的bit位。

• initial:
  • res=1
• bit[0]=1:
  • res=res*x
  • X=x^2
• bit[1]=0:
  • res
  • X=(x^2)^2=x^(2^2)=x^4
• bit[2]=0:
  • res
  • X=(x^4)^2=x^(2^3)=x^8
• bit[3]=1:
  • res=res*x
  • X=(x^8)^2=x^(2^4)=x^16

代码参考：

class Solution {
public:
    double myPow(double x, int n) {
        double res=1;
        long pn = abs(n);
        while(pn>0) {
            if(pn&1) res*=x;
            pn = pn>>1;
            x*=x;
        }
        return n>=0?res:1/res;
    }
};