/**
* 编写一种递归方法，它返回数N的二进制中表示1的个数。利用这样一个事实：N为奇数，其1的个数为N/2的二进制中1的个数加1.
* @author wulei
*
*/
public class BinaryTest {
//Main method.
public static void main(String[] args) {
//For example ,make N equals 11 ,the result shows 2 (1011)
System.out.println(numOfEven(11));
//For example ,make N equals 512 ,the result shows 1 (1000000000)
System.out.println(numOfEven(512));
}
//The static method of numOfEven is the recursive method.
public static int numOfEven(int x) {
//递归必须要有基准，不然不会结束推进
if(x==0){
return 0;
}else if(x%2 != 0){
return numOfEven(x>>1) + 1;
}else { //可以用while减少遍历次数，也可以直接多几次遍历
/* while (x%2==0) {
x=x>>1;
}*/
return numOfEven(x>>1);
}
}
}
参考：http://www.vimiy.com/a/egao/62709.html