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.
For num = 5 you should return [0,1,1,2,1,2].
- 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.
- 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?
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。
dp[i] = dp[i >> 1] + (i & 1)