counting number of ones in binary representation of a number

Possible Duplicate:
Best algorithm to count the number of set bits in a 32-bit integer?

I want to find out how many 1s are there in binary representation of a number.I have 2 logic .

  int count =0;
int no = 4;

while(no!=0){
    int d = no%2;
    if(d==1)
        count++;
    no = no/2;
    str = str+ d;
}

Use Java API(java 5 or above).

Integer.bitCount(int);
Long.bitCount(long);

NOTE: The above java methods are based on hacker's delight

比以前的答案更快：（与1比特的数量成比例，而不是总比特数量）

public class Foo {
  public static void main(String[] argv) throws Exception {
    int no = 12345;
    int count;
    for (count = 0; no > 0; ++count) {
      no &= no - 1;
    }
    System.out.println(count);
  }
}

Looks like c/c++/c#, if so you have shifting.. just loop to N-1 bits from 0 and use sum+=(value>>i)&1

Ie: you always check the last/right most bit but move the binary representation of the number to the right for every iteration until you have no more bits to check.

Also, think about signed/unsigned and any integer format. But you dont state how that should be handled in the question.

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