Content

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++)?

Solution

1. 位运算

Java

class Solution {
    public int[] countBits(int n) {
        // 0 <= n <= 10⁵
        int[] ans = new int[n + 1];
        for (int i = 0; i <= n; i++) {
            int j = i;
            while (j > 0) {
                ans[i] += j & 1;
                j >>= 1;
            }
        }
        return ans;
    }
}

2. 动态规划

Java

class Solution {
    public int[] countBits(int n) {
        // 0 <= n <= 10⁵
        int[] ans = new int[n + 1];
        for (int i = 0; i <= n; i++) {
            ans[i] = (i & 1) + ans[i >> 1];
        }
        return ans;
    }
}