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java中实现递归计算二进制表示中1的个数

投稿:hebedich

这是一个很有意思的问题,是在面试中特别容易被问到的问题之一,解决这个问题第一想法肯定是一位一位的去判断,是1计数器+1,否则不操作,跳到下一位,十分容易,编程初学者就可以做得到!

借助Java语言,运用递归算法计算整数N的二进制表示中1的个数

/*use the recursive algorithme to calculate 
 * the number of "1" in the binary expression
 * of an Integer N.
 * Note:if N is an odd, then
 * the result is the result of N/2 plus 1.
 * And the program use the bit operation to
 * improve efficency ,though it's seemingly
 * not necessary ,but the idea I think is good.
 * The program is writed by Zewang Zhang ,at
 * 2015-5-4,in SYSU dorms.
 */
 
public class CalculateNumberInBinaryExpression {
  //Main method.
  public static void main(String[] args) {
     
    //For example ,make N equals 13 ,the result shows 3
    System.out.println(numOfEven(13));
     
    //For example ,make N equals 128 ,the result shows 1
    System.out.println(numOfEven(128));
  }
   
  //The static method of numOfEven is the recursive method.
  public static int numOfEven(int x) {
     
    //The base of recursive.
    if(x==0) {
      return 0;
    }
     
    //If x is an odd.
    else if(x%2!=0) {
      return numOfEven(x>>1)+1;
    }
     
    //If x is an even except 0.
    else {
      while(x%2==0) {
        x=(x>>1);
      }
      return numOfEven(x);
    }
  }
}

来个最简单的,不过未测试:)

public int a(int i){
    if(i==0||i==1) return i;
    return i%2+a(i/2);

}

以上所述就是本文的全部内容了,希望大家能够喜欢。

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