# 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 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.
# Solution
Approach 1: Identify the rule.
# Code (Python)
Approach 1:
# Code (C++)
Approach 1:
class Solution {
public:
vector<int> countBits(int num) {
vector<int> res;
if (num >= 0) res.push_back(0);
if (num >= 1) res.push_back(1);
int base = 2;
int i = 0;
for (int n = 2; n <= num; ++n)
{
if (i == base)
{
i = 0;
base *= 2;
}
res.push_back(res[i] + 1);
i++;
}
return res;
}
};
class Solution {
public:
vector<int> countBits(int num) {
vector<int> res;
res.push_back(0);
for (int i = 1; i <= num; ++i) {
// option 1
// int j = i - (i & -i); // or int j = i & (i - 1);
// res.push_back(res[j] + 1);
// option 2
res.push_back(res[i>>1] + (i & 1));
}
return res;
}
};