1、逐位法

public static int bitCount(int n){
    int count = 0;
    while(n != 0){
        count += n & 1;
        n >>>= 1; // 无符号右移
    }
    return count;
}

2、查表法

private static final int[] oneNumberTable ={
    0,1,1,2,1,2,2,3,1,2,2,3,2,3,3,4,1,2,2,3,2,3,3,4,2,3,3,4,3,4,4,5,
    1,2,2,3,2,3,3,4,2,3,3,4,3,4,4,5,2,3,3,4,3,4,4,5,3,4,4,5,4,5,5,6,
    1,2,2,3,2,3,3,4,2,3,3,4,3,4,4,5,2,3,3,4,3,4,4,5,3,4,4,5,4,5,5,6,
    2,3,3,4,3,4,4,5,3,4,4,5,4,5,5,6,3,4,4,5,4,5,5,6,4,5,5,6,5,6,6,7,
    1,2,2,3,2,3,3,4,2,3,3,4,3,4,4,5,2,3,3,4,3,4,4,5,3,4,4,5,4,5,5,6,
    2,3,3,4,3,4,4,5,3,4,4,5,4,5,5,6,3,4,4,5,4,5,5,6,4,5,5,6,5,6,6,7,
    2,3,3,4,3,4,4,5,3,4,4,5,4,5,5,6,3,4,4,5,4,5,5,6,4,5,5,6,5,6,6,7,
    3,4,4,5,4,5,5,6,4,5,5,6,5,6,6,7,4,5,5,6,5,6,6,7,5,6,6,7,6,7,7,8
};
public static int bitCount(int n){
    int count = 0;
    for (int i = 0; i < 32; i += 8){
        count += oneNumberTable[(n >> i) & 0xFF];
    }
    return count;
}

3、Brian Kernighan

public static int bitCount(int n){
    int count = 0;
    while(n != 0){
        n = n & (n - 1);
        count++;
    }
    return count;
}

4、Hamming Weight

public static int bitCount(int n){
    n = (n & 0x55555555) + ((n >>> 1) & 0x55555555);
    n = (n & 0x33333333) + ((n >>> 2) & 0x33333333);
    n = (n & 0x0F0F0F0F) + ((n >>> 4) & 0x0F0F0F0F);
    n = (n & 0x00FF00FF) + ((n >>> 8) & 0x00FF00FF);
    n = (n & 0x0000FFFF) + ((n >>> 16) & 0x0000FFFF);
    return n;
}

Java计算二进制数中1的个数