338. Counting Bits

📋 Đề Bài

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

Follow up:

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++)?

💻 Lời Giải

C++ 0338-counting-bits.cpp

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