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].

要求O(n)时间

巧妙的利用了i&(i - 1)， 这个本来是用来判断一个数是否是2的指数的快捷方法，比如8，二进制位1000, 那么8&(8-1)为0，只要为0就是2的指数, 那么我们现在来看一下0到15的数字和其对应的i&(i - 1)值：

i    bin       '1'    i&(i-1)   0000    0-----------------------   0001    1    0000-----------------------   0010    1    0000   0011    2    0010-----------------------   0100    1    0000   0101    2    0100   0110    2    0100   0111    3    0110-----------------------   1000    1    0000   1001    2    1000   1010    2    1000   1011    3    1010   1100    2    1000   1101    3    1100   1110    3    1100   1111    4    1110

class Solution {public:    vector<int> countBits(int num) {        vector<int> res(num + 1, 0);        for (int i = 1; i <= num; ++i) {            res[i] = res[i & (i - 1)] + 1;        }        return res;    }};