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338. Counting Bits

题目

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:

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

Example 2:

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.

题目大意

给定一个非负整数 num。对于 0 ≤ i ≤ num 范围中的每个数字 i ，计算其二进制数中的 1 的数目并将它们作为数组返回。

解题思路

给出一个数，要求计算出 0 ≤ i ≤ num 中每个数的二进制位 1 的个数。

这一题就是利用二进制位运算的经典题。

  X&1==1or==0，可以用 X&1 判断奇偶性，X&1>0 即奇数。
  X = X & (X-1) 清零最低位的1
  X & -X => 得到最低位的1 
  X&~X=>0

Uh oh!

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