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Write a function that takes an unsigned integer and returns the number of ’1‘ bits it has (also known as the Hamming weight).

For example, the 32-bit integer ’11‘ has binary representation 00000000000000000000000000001011, so the function should return 3.

Credits:

Special thanks to @ts for adding this problem and creating all test cases.

思路分析：这题考察位运算，比较容易想到的思路是不断进行左移位操作然后并且和1做&运算，就可以每次判定最后一位是否是1，从而得到1的个数，见注释的code,这种算法需要做32次移位操作。这题有更快的解法，那就是利用n&(n-1)可以去掉最右侧的1，例如

n       000110100

n-1    000110011

&后   000110000

最右侧的1消失。于是我们可以不断做这样的操作直到n变成0。就可以知道有多少个1。n&(n-1)还有一个应用，就是快速判断某个数是否是2的k次方，即判断某个数的二进制表示是否只包括一个1，

AC Code

public class Solution {
    // you need to treat n as an unsigned value
    public int hammingWeight(int n) {
        int counter = 0;
        /*while(n!=0){
            n = n & (n-1);
            counter++;
        }
        return counter;*/
        for(int i = 0; i<32 ; i++){
            counter += n&1;
            n = n>> 1;
        }
        return counter;
    }
}

int x = 1;
String bin = Integer.toBinaryString(x);
System.Out.Println(bin);

上面给出了把某个整数转化成其二进制表示的方法，代码输出0000 0001

问题

I am very confused on right shift operation on negative number, here is the code.

int n = -15;
System.out.println(Integer.toBinaryString(n));
int mask = n >> 31;
System.out.println(Integer.toBinaryString(mask));

And the result is:

11111111111111111111111111110001
11111111111111111111111111111111

Why right shifting a negative number by 31 not 1 (the sign bit)?

Answer

Because in Java there are no unsigned datatypes, there are two types of right shifts: arithmetic shift >> and logical shift >>>. http://docs.oracle.com/javase/tutorial/java/nutsandbolts/op3.html

Arithmetic shift >> will keep the sign bit.

Unsigned shift >>> will not keep the sign bit (thus filling 0s).

(images from Wikipedia)

LeetCode Number of 1 Bits

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原文地址：http://blog.csdn.net/yangliuy/article/details/46502357