问题描述：
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.
Credits:
Special thanks to @ syedee for adding this problem and creating all test cases.
解题思路：
我们可以先观察一下数字的二进制形式：
0 --------->0
1 ---------> 1
2 ---------> 10
3 ---------> 11
4 ---------> 100
5---------->1001
二进制为逢二进一，当当前数字全部为1时，再往前一位变为1.
其实观察上面的形式我们可以发现：
当进位后，重复进位前的过程直至再次进一产生新的1。
所以我们可以不断循环当前的ret数组，对ret数组内的值+1后然后加入ret数组尾部。
这里的count代表的时即将计算的数字，所以count == num时，num的值还没有加入
当count > num时，此时已经加入。
时间复杂度为O(n)
空间复杂度为O(n)
代码：
class Solution { public: vector<int> countBits(int num) { vector<int> ret; ret.push_back(0); if(num == 0) return ret; ret.push_back(1); int count = 2; while(count <= num){ int last_end = ret.size(); for(int j = 0; j < last_end; j++){ ret.push_back(ret[j] + 1); count++; if(count > num) break; } } return ret; } };