https://leetcode.com/problems/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 1:
Input: 2
Output: [0,1,1]
Example 2:
Input: 5
Output: [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.
对于每个数，获得他右移之后得到的数的 1 的个数，然后求奇偶看其最后一位是 1 还是 0 即可。
class Solution {
public:
vector<int> countBits(int num) {
vector<int> d;
d.push_back(0);
for (int i = 1; i <= num; i++) {
int nd = d[i>>1] + i%2;
d.push_back(nd);
}
return d;
}
};
