[LeetCode] Counting Bits

最新推荐文章于 2025-09-12 18:45:12 发布

转载最新推荐文章于 2025-09-12 18:45:12 发布 · 51 阅读

0 ·

CC 4.0 BY-SA版权

原文链接：https://segmentfault.com/a/1190000005604740

文章标签：

#c/c++

本文介绍了一种高效算法，用于计算从0到给定整数范围内每个数的二进制表示中1的个数。通过使用位运算技巧，该算法实现了线性时间复杂度O(n)内的解决方案。

Problem

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

For num = 5 you should return [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.

Hint

You should make use of what you have produced already.
Divide the numbers in ranges like [2-3], [4-7], [8-15] and so on. And try to generate new range from previous.
Or does the odd/even status of the number help you in calculating the number of 1s?

Note

应用公式f[i] = f[i/2] + (i%2);
并优化此公式为f[i] = f[i>>2] + (i&1)，减少计算时间。

Solution

public class Solution {
    public int[] countBits(int num) {
        int[] dp = new int[num+1];
        for (int i = 1; i <= num; i++) dp[i] = dp[i>>1] + (i&1);
        return dp;
    }
}