leetcode之Counting_Bits

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.
Credits:
Special thanks to @ syedee for adding this problem and creating all test cases.

题目如上，就是给你一个数字，让给输出从0到这个数字之间所有数字转化为二进制后1的个数。

// 0    0000    0                
// 1    0001    1   
// 2    0010    1
// 3    0011    2
// 4    0100    1
// 5    0101    2
// 6    0110    2
// 7    0111    3
// -------------        
// 8    1000    1        
// 9    1001    2        
// 10   1010    2        
// 11   1011    3        
// 12   1100    2         
// 13   1101    3
// 14   1110    3
// 15   1111    4
// 16   10000   1
// 17   10001   2

// 31    1 1111 5
// 63   11 1111 6

解法一：

var countBits = function(num) {
      debugger
     var res = [0]
      for (var i = 1; i <= num; ++i) {
          if (i % 2 == 0) res.push(res[parseInt(i / 2)])
          else res.push(res[parseInt((i / 2))] + 1)
      }
      return res

解题思路：
从1开始，遇到偶数时，其1的个数和该偶数除以2得到的数字的1的个数相同，
遇到奇数时，其1的个数等于该奇数除以2得到的数字的1的个数然后再加上1
第二种思路：

var countBits = function(num) {
      debugger
     var res = [0]
        for (int i = 1; i <= num; ++i) {
            res[i] = res[i & (i - 1)] + 1;
        }
      return res

解题思路：每个i值都是i&(i-1)对应的值加1 （简单暴力，神奇的n与n-1）

---

第二种参考自Grandyangde的博客:

---