高效求a的n次幂的算法

public class A的N次幂 {

	public static void main(String[] args) {
		int a = 2;
		int n = 60;
		long t = System.nanoTime();   // 纳秒
		System.out.println(pow0(a, n));
		System.out.println("pow0()所花时间："+(System.nanoTime()-t)+"ns");
		
		t = System.nanoTime();
		System.out.println(pow(a, n));
		System.out.println("pow()所花时间："+(System.nanoTime()-t)+"ns");

	}
	
	// O(N)
	private static long pow0(int a,int n){
		long res = 1;
		for(int i=0;i<n;i++){
			res *= a;
		}
		return res;
	}
	
	private static long pow(int a,int n){
		if (n==0) {
			return 1;
		}
		long res = a;
		int ex = 1;
		while((ex<<1)<=n){
			res = res * res;
			ex <<=1;
		}
		return res*pow(a, n-ex);
	}

}