# 338.counting-bits

## Statement

• Difficulty: Easy
• Tag: `位运算` `动态规划`

``````输入：n = 2

0 –> 0
1 –> 1
2 –> 10
``````

``````输入：n = 5

0 –> 0
1 –> 1
2 –> 10
3 –> 11
4 –> 100
5 –> 101
``````

• `0 <= n <= 105`

• 很容易就能实现时间复杂度为 `O(n log n)` 的解决方案，你可以在线性时间复杂度 `O(n)` 内用一趟扫描解决此问题吗？
• 你能不使用任何内置函数解决此问题吗？（如，C++ 中的 `__builtin_popcount`

• Difficulty: Easy
• Tag: `Bit Manipulation` `Dynamic Programming`

Given an integer `n`, return an array `ans` of length `n + 1` such that for each `i` (`0 <= i <= n`), `ans[i]` is the number of `1`'s in the binary representation of `i`.

Example 1:

``````Input: n = 2
Output: [0,1,1]
Explanation:
0 –> 0
1 –> 1
2 –> 10
``````

Example 2:

``````Input: n = 5
Output: [0,1,1,2,1,2]
Explanation:
0 –> 0
1 –> 1
2 –> 10
3 –> 11
4 –> 100
5 –> 101
``````

Constraints:

• `0 <= n <= 105`

• It is very easy to come up with a solution with a runtime of `O(n log n)`. Can you do it in linear time `O(n)` and possibly in a single pass?
• Can you do it without using any built-in function (i.e., like `__builtin_popcount` in C++)?

## Solution

``````#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>

#define endl "\n"
#define fi first
#define se second
#define all(x) begin(x), end(x)
#define rall rbegin(a), rend(a)
#define bitcnt(x) (__builtin_popcountll(x))
#define complete_unique(a) a.erase(unique(begin(a), end(a)), end(a))
#define mst(x, a) memset(x, a, sizeof(x))
#define MP make_pair

using ll = long long;
using ull = unsigned long long;
using db = double;
using ld = long double;
using VLL = std::vector<ll>;
using VI = std::vector<int>;
using PII = std::pair<int, int>;
using PLL = std::pair<ll, ll>;

using namespace __gnu_pbds;
using namespace std;
template <typename T>
using ordered_set = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
const ll mod = 1e9 + 7;

template <typename T, typename S>
inline bool chmax(T &a, const S &b) {
return a < b ? a = b, 1 : 0;
}

template <typename T, typename S>
inline bool chmin(T &a, const S &b) {
return a > b ? a = b, 1 : 0;
}

#ifdef LOCAL
#include <debug.hpp>
#else
#define dbg(...)
#endif

class Solution {
public:
vector<int> countBits(int n) {
auto res = vector<int>();
for (int i = 0; i <= n; i++) {
res.push_back(__builtin_popcount(i));
}

return res;
}
};

#ifdef LOCAL

int main() {
return 0;
}

#endif
``````