leetCode #338 - Counting Bits

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本文介绍了一种算法，该算法可以计算从1到n范围内每个整数在二进制表示下1的个数，并返回这些计数构成的数组。该算法的时间复杂度为O(n * sizeof(integer))，空间复杂度为O(n)。

题目描述：

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.

求1~n每个数的为1 的位数，时间复杂的为O(n*sizeof(integer))，空间复杂度为O(n)。

class Solution {
public:
    vector<int> countBits(int num) {
        vector<int> bits_num(num+1,0);
        for(int i=0;i<=num;i++)
            bits_num[i]=get_bits_num(i);
        return bits_num;
    }
    
    int get_bits_num(int num)
    {
        int count=0;
        while(num>0)
        {
            if(num%2==1) count++;
            num/=2;
        }
        return count;
    }
};