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.

这道题要求充分利用前面的结果，我们可以找到以下规律：
如果一个数是2的幂次，那么它1的个数就是1个
如果不是，那这个数可以拆成比他小的那个最大的2的幂次的和另一个数的和，1的个数也是这两个数中1的个数的和。其中2的幂次的1的个数永远为1，另一个数的1的个数已经计算过了。
比如14这个数，其二进制码为1110，拆成8+6，1000+0110。

/**
 * @param {number} num
 * @return {number[]}
 */
var countBits = function(num) {
    var res=[];
    res[0]=0;
    var k=0;
    for(var i=1;i<=num;i++){
        if((i&(i-1))===0){
            k=i;
            res[i]=1; 
        }
        else {
            res[i]=1+res[i-k];
        }
    }
    return res;
};