338. Counting Bits

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].

Follow up:

• It is very easy to come up with a solution with run time O(n*sizeof(integer)). But can you do it in linear time O(n) /possibly in a single pass?
• Space complexity should be O(n).
• Can you do it like a boss? Do it without using any builtin function like __builtin_popcount in c++ or in any other language.

题目含义：统计二进制数字num以内每个数中‘1’的个数

 

    
    int countbit(int i)
    {
        //无关trick   i&(i - 1)， 这个本来是用来判断一个数是否是2的指数的快捷方法，比如8，二进制位1000, 那么8&(8-1)为0，只要为0就是2的指数
//        按照定义做， x&(x-1)可以消去最右边的1
        int count = 0;
        while(i>0)
        {
            i &= (i-1);
            count ++;
        }
        return count;
    }
    
    public int[] countBits(int num) {
        int[] res = new int[num+1];
        if (num == 0) return res;
        res[0] = 0;
        for (int i=1;i<=num;i++)
        {
//            res[i] = countbit(i);//解法一
            res[i] = res[i & (i - 1)] + 1;//每个i值都是i&(i-1)对应的值加1
        }
        return res;       
    }