# Leetcode之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.

``````class Solution {
public:
vector<int> countBits(int num) {
vector<int> res = { 0 };
if (num == 0)return res;
int mark = 1;
for (int i = 1; i <= num; i++) {
if (i == mark) {
res.push_back(1);
mark *= 2;
}
else {
int temp = 1 + res[i - mark/2];
res.push_back(temp);
}
}
return res;
}
};``````

``````class Solution {
public:
vector<int> countBits(int num) {
vector<int> res{0};
for (int i = 1; i <= num; ++i) {
if (i % 2 == 0) res.push_back(res[i / 2]);
else res.push_back(res[i / 2] + 1);
}
return res;
}
};``````

``````class Solution {
public:
vector<int> countBits(int num) {
vector<int> res(num + 1, 0);
for (int i = 1; i <= num; ++i) {
res[i] = res[i & (i - 1)] + 1;
}
return res;
}
};``````