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.

Hint:

  • You should make use of what you have produced already.
  • Divide the numbers in ranges like [2-3], [4-7], [8-15] and so on. And try to generate new range from previous.
  • Or does the odd/even status of the number help you in calculating the number of 1s?

找规律

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2
3
4
5
6
7
8
9
0	    0
1	    1
2	   10
3	   11
4	  100
5	  101
6	  110
7	  111
...

每一个数可以分成两部分，最后一位(the least significant), 和其他的所有位。譬如说7 (111), 由11和1组成。所以7中1的数量为11和1中1的数量和。
可得dp[i] = dp[i >> 1] + (i & 1)

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public int[] countBits(int num) {
    int[] dp = new int[num + 1];
    // dp[0] = 0;
    for (int i = 1; i <= num; i++) {
        dp[i] = dp[i >> 1] + (i & 1);
    }
    return dp;
}