Eighty. Find the number of 1s in binary (bit operation)

Please implement a function, input an integer, and output the number of 1s in the binary representation of the number.

Method 1: Move 1 (Move to the left)

import java.util.Scanner;

public class LianXi {
    
    
	public static void main(String[] args){
    
    
		Scanner in = new Scanner(System.in);
		int N = in.nextInt();		
		
		//二进制转化
	    System.out.println(Integer.toString(N,2));
		
		int count = 0;		
		//因为整数有32位，所以进行32次循环
		for(int i = 0; i < 32; i++){
    
    
			/*将二进制转化后的整数与1左移i为得到的数进行与运算，
			     如果等于1左移i为得到的数，说明有1。*/
			if((N&(1<<i)) == (1<<i)){
    
    
				count++;
		    }
		}
		System.out.println(count);
	}
}

Method 2: Shift the binary number (shift to the right)

import java.util.Scanner;

public class LianXi {
    
    
	public static void main(String[] args){
    
    
		Scanner in = new Scanner(System.in);
		int N = in.nextInt();		
		//二进制转化
	    System.out.println(Integer.toString(N,2));
		
		int count = 0;
		
		//因为整数有32位，所以进行32次循环
		for(int i = 0; i < 32; i++){
    
    
			/*将二进制转化后的整数向右挪i位和1进行与运算，
			     如果等于1，说明有1。*/
			if(((N>>i)&1) == 1){
    
    
				count++;
		    }
		}
		System.out.println(count);
	}
}

Method three: subtraction operation (recursively subtract 1 from a binary number)

import java.util.Scanner;

public class LianXi {
    
    
	public static void main(String[] args){
    
    
		Scanner in = new Scanner(System.in);
		int N = in.nextInt();		
		//二进制转化
	    System.out.println(Integer.toString(N,2));
		
		int count = 0;
		/* 转化后的二进制数减1和自身做与运算，将会消掉最低位的1。
		       以此往复最终将二进制数变成0，并以此为终止条件。
		*/
		while(N!=0){
    
    
			N = ((N-1)&N);
			count++;
		}
		System.out.println(count);
	}
}