338. Counting Bits 和191. Number of 1 Bits

第一题、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:
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.
方法一、i&(i-1)可以消除最后一个1的特性，代码如下：

vector<int> countBits(int num) {
        vector<int> ret(num+1,0);
        for(int i = 1; i <= num; i++)
            ret[i] = ret[i&(i-1)] + 1; //由于i&(i-1)可以消除最后一个1，所以可以这样做
        return ret;
    }

方法二、普通的做法:

vector<int> countBits(int num) {
        vector<int> total(num+1);
        for(int i=0;i<=num;i++)
        {
            total[i]=0;
            int j=i;
            while(j)
            {
                j&=(j-1);
                total[i]++;
            }
        }

        return total;
    }

第二题、191. Number of 1 Bits
Write a function that takes an unsigned integer and returns the number of ’1’ bits it has (also known as the Hamming weight).

For example, the 32-bit integer ’11’ has binary representation 00000000000000000000000000001011, so the function should return 3
题意：统计一个整数的32位的二进制中1的个数，和上题的思路类似

int hammingWeight(uint32_t n) {
        int count = 0;
        while(n){
            n &= (n -1);
            count++;
        }

        return count;
    }