LeetCode – Counting Bits (Java)

Given a non negative integer number num. For every numbers i in the range 0 ≤ i ≤ num calculate the number of 1's in their binary representation and return them as an array.

Example:

For num = 5 you should return [0,1,1,2,1,2].

1. Naive Solution

We can simply count bits for each number like the following:

public int[] countBits(int num) {
    int[] result = new int[num+1];
 
    for(int i=0; i<=num; i++){
        result[i] = countEach(i);
    }
 
    return result;
}
 
public int countEach(int num){
    int result = 0;
 
    while(num!=0){
        if(num%2==1){
            result++;
        }
        num = num/2;
    }
 
    return result;
}

2. Improved Solution

For number 2(10), 4(100), 8(1000), 16(10000), ..., the number of 1's is 1. Any other number can be converted to be 2^m + x. For example, 9=8+1, 10=8+2. The number of 1's for any other number is 1 + # of 1's in x.

Counting Bits (Java)

public int[] countBits(int num) {
    int[] result = new int[num+1];
 
    int p = 1; //p tracks the index for number x
    int pow = 1;
    for(int i=1; i<=num; i++){
        if(i==pow){
            result[i] = 1;
            pow <<= 1;
            p = 1;
        }else{
            result[i] = result[p]+1;
            p++;
        }
 
    }
 
    return result;
}
Category >> Algorithms  
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