（LeetCode）Counting Bits -- 求二进制中1的个数

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.

解题分析：

 方法一：

创建一个集合，把每个数的1的个数求出来，然后存放到位置中。

__author__ = 'jiuzhang'# -*- coding:utf-8 -*-class Solution(object):    def countBits(self, num):        if num == 0:            return [0]        res = [0]        bit = 0        tmp = num        while tmp > 0:            tmp /= 2            bit += 1        for i in xrange(bit - 1):            t_res = []            for j in res:                t_res.append(j + 1)            res = res + t_res        for i in res[:(num - pow(2, bit - 1)) + 1]:            res.append(i + 1)        return res

方法二：

        利用位运算，右移，并做与运算。

__author__ = 'jiuzhang'class Solution(object):    def countBits(self, num):        """        :type num: int        :rtype: List[int]        """        ans = [0]        for x in range(1, num + 1):            ans += ans[x >> 1] + (x & 1),        return ans