338.LeetCode Counting Bits(medium)[二进制位处理]

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?
这个题目的意思是要求num范围中每个数字的二进制中1的个数，常规思路是对每个数字除以2然后得到二进制位中1的个数的和，但是由于题目要求效率要是O(n)，那么就要求在常规时间里面算出当前数字中1的个数。这里采用的思路是这样的，如果是2的幂次那么肯定1的个数是1，然后计算当前num和2的幂次的中间的插值，找到这个插值数的1的个数加上1就得到当前的num的1的个数了。

class Solution {
public:
vector<int> countBits(int num) {
vector<int> result;
//if(num == 0)
result.push_back(0);
//result.push_back(1);
//result.push_back(1);
//result.push_back(2);
//if(num<=3) return result;
int flag = 0;
for(int i = 1;i<=num;i++)
{
if((i&(i-1)) == 0)//此时为2的整次幂，只有1个1
{
flag = i;
result.push_back(1);
}else
{
int step = i - flag;
result.push_back(1+ result[step]);
}
}
return result;
}
};