leetcode-338 Counting Bits

338. Counting Bits

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 1:

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Input: 2
Output: [0,1,1]

Example 2:

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Input: 5
Output: [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.

Analyse

给定一个非负整数num，计算[0, num]范围内所有数的二进制表示中 1 的个数，以数组的形式输出

先解决怎么数二进制中的 1 的问题，二进制的 1 除了最后一位，都代表 2 的幂次，
比如 7 的二进制111中的 1 分别代表，$2^2$， $2^1$，$2^0$，

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int CountBit(int num)
{
    int count = 0;
    while (num != 0)
    {
        if (num % 2 == 1)
        {
            count++;
        }
        num = num >> 1;
    }
    return count;
}

vector<int> countBits(int num)
{
    vector<int> vec;
    for(int i = 0; i <= num; i++)
    {
        vec.push_back(CountBit(i));
    }
    return vec;
}