leetcode-338-Counting Bits

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.

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

给定num，对于每个 0<= i <=num ， 计算i的2进制中1的个数。

若对每个数，进行每位的判断，时间复杂度是O(n*logn)。

可以从每个数的最低位开始分析，例如1001001 ，它的二进制1的个数等于100100 总二进制个数 + 1 。

即 f = f/2 + (f 的最低位)

class Solution {
public:
vector<int> countBits(int num) {
vector<int> v(num + 1 , 0) ;
for (int i = 1 ; i <= num ; i ++) {
v[i] = v[i>>1] + i%2 ;
}
return v ;
}
};