338. Counting Bits

最新推荐文章于 2024-02-16 08:15:42 发布

原创最新推荐文章于 2024-02-16 08:15:42 发布 · 249 阅读

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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，并以数组形式返回这些计数值。此算法通过巧妙利用位操作实现了线性时间复杂度。

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.

#include <iostream>
#include <vector>
using namespace std;
class Solution {
public:
vector<int> countBits(int num) {
vector<int> r;
r.push_back(0);
for(int i=1;i<=num;i++)
{
int re = r[i >> 1] + (i & 1);//根据规律，若num的二进制表示最后一位为0，则它的1个数等于它除以2后得到的数中的1的个数，若最后一位为1，则需要再加1
r.push_back(re);
}
return r;
}
};