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PRoblem
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?
Note
应用公式f[i] = f[i/2] + (i%2);
并优化此公式为f[i] = f[i>>2] + (i&1)，减少计算时间。
Solution
public class Solution {
public int[] countBits(int num) {
int[] dp = new int[num+1];
for (int i = 1; i <= num; i++) dp[i] = dp[i>>1] + (i&1);
return dp;
}
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