number of 1's in 32 bit number

问题

I am lookin for a method to have number of 1's in 32 bit number without using a loop in between. can any body help me and provide me the code or algorithm to do so. Thanks in advance.

回答1:

See Integer.bitCount(int). You can refer to the source code if you want to see how it works; many of the Integer class's bit twiddling routines are taken from Hacker's Delight.

回答2:

See the canonical reference: Bit Twiddling Hacks

回答3:

Short, obscenely optimized answer (in C):

int pop(unsigned x) {
   x = x - ((x >> 1) & 0x55555555);
   x = (x & 0x33333333) + ((x >> 2) & 0x33333333);
   x = (x + (x >> 4)) & 0x0F0F0F0F;
   x = x + (x >> 8);
   x = x + (x >> 16);
   return x & 0x0000003F;
}

To see why this magic works, see The Quest for an Accelerated Population Count by Henry S. Warren, Jr. chapter 10 in Beautiful Code.

回答4:

Split the 32 bit number into four 8 bit numbers (see bit shifting operator, casting etc.)

Then use a lookup with 256 entries that converts the 8 bit number into a count of bits set. Add the four results, presto!

Also, see what Mitch Wheat said - bit fiddling can be a lot of fun ;)

回答5:

You can define it recursively:

int bitcount(int x) {
  return (x==0) ? 0 : (x & 1 + bitcount(x/2));
}

The code above is not tested, and probably only works for x>=0. Hopefully, you will get the idea anyways...

回答6:

My personal favourite, directly from Bit Twiddling Hacks:

v = v - ((v >> 1) & 0x55555555);
v = (v & 0x33333333) + ((v >> 2) & 0x33333333);
c = ((v + (v >> 4) & 0xF0F0F0F) * 0x1010101) >> 24;

回答7:

Following is JDK 1.5 implementation of of Integer.bitCount

public static int bitCount(int i) {
    // HD, Figure 5-2
i = i - ((i >>> 1) & 0x55555555);
i = (i & 0x33333333) + ((i >>> 2) & 0x33333333);
i = (i + (i >>> 4)) & 0x0f0f0f0f;
i = i + (i >>> 8);
i = i + (i >>> 16);
return i & 0x3f;
}

来源：https://stackoverflow.com/questions/1458314/number-of-1s-in-32-bit-number

标签

java

algorithm

bit-manipulation