java中实现递归计算二进制表示中1的个数
发布时间:2020-12-14 17:45:26 所属栏目:Java 来源:网络整理
导读: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
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。 (编辑:李大同) 【声明】本站内容均来自网络,其相关言论仅代表作者个人观点,不代表本站立场。若无意侵犯到您的权利,请及时与联系站长删除相关内容! |