Leetcode 191. number of 1 bits bit operation /easy

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1.Description

Write a function , Input is an unsigned integer （ In the form of a binary string ）, Returns the number of digits in its binary expression as ‘1’ The number of （ Also known as Han Ming weight ）.

Tips ：

Please note that , In some languages （ Such as Java） in , There is no unsigned integer type . under these circumstances , Both input and output will be specified as signed integer types , And should not affect your implementation , Because whether integers are signed or unsigned , Its internal binary representation is the same .
stay Java in , The compiler uses binary complement notation to represent signed integers . therefore , Above Example 3 in , The input represents a signed integer -3.

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2.Example

 Input ：00000000000000000000000000001011
 Output ：3
 explain ： Binary string of input  00000000000000000000000000001011  in , There are three of them  '1'.

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3.Solution

1. Right shift to position

You can move right 32 Times or move right to n be equal to 0 until .
Be careful ：Java Medium “>>” yes Count right , Abandon the lowest position , The high bits are filled with sign bits ;
“>>>” Logical shift right ： Abandon the lowest position , High level 0 fill .
Therefore, when using the second method, you need to use “>>>”.

public class Solution {
    
    // you need to treat n as an unsigned value
    public int hammingWeight(int n) {
    
        int count = 0;
    	while(n!=0) {
    
        	if((n&1)==1) {
    
        		count++;
        	}
            n=n>>>1;
        }
    	return count;
    }
}

2. Bit operation optimization

Use n = n&(n-1) You can get n The lowest position of 1 become 0 The number after , Therefore, as long as the operation is carried out circularly, it is enough to count the total number of operations .

public class Solution {
    
    public int hammingWeight(int n) {
    
        int ret = 0;
        while (n != 0) {
    
            n &= n - 1;
            ret++;
        }
        return ret;
    }
}