50. Pow(x, n)

最开是在剑指offer上面看到这道题，今天遇到试了一下发现不可以，上网找了一下才发现还有更巧妙的方法。

Consider the binary representation of n. For example, if it is "10001011", then x^n = x^(1+2+8+128) = x^1 * x^2 * x^8 * x^128. Thus, we don't want to loop n times to calculate x^n. To speed up, we loop through each bit, if the i-th bit is 1, then we add x^(1 << i) to the result. Since (1 << i) is a power of 2, x^(1<<(i+1)) = square(x^(1<<i)). The loop executes for a maximum of log(n) times.

比较简单就不翻译了，下面是我的实现代码

public static double myPow(double x, int n) {
        if(n<0){
        	if(n==Integer.MIN_VALUE){
        		return 1.0/myPow(x,Integer.MAX_VALUE)*x;
        	}else{
        		return 1.0/myPow(x, -n);
        	}
        }
        if(n==0){
        	return 1.0;
        }
        double re=1.0;
        for(; n>0; x*=x, n>>=1){
        	if((n&1)>0){
        		re*=x;
        	}
        }
        return re;
    }