# 338. Counting Bits 比特位计数

@TOC

## # 题目描述

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]`.

• 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.

## # 解题方法

``````   0000    0
-------------
0001    1
-------------
0010    1
0011    2
-------------
0100    1
0101    2
0110    2
0111    3
-------------
1000    1
1001    2
1010    2
1011    3
1100    2
1101    3
1110    3
1111    4
``````

Python代码如下：

``````class Solution(object):
def countBits(self, num):
"""
:type num: int
:rtype: List[int]
"""
res = [0] * (num + 1)
for i in range(1, num + 1):
res[i] = res[i / 2] + i % 2
return res
``````

Java代码如下：

``````public class Solution {
public int[] countBits(int num) {
for(int i = 1; i < answer.length; i++){
}
}
}
``````

C++代码如下：

``````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 / 2] + i % 2;
}
return res;
}
};
``````

## # 日期

2017 年 4 月 25 日 2018 年 12 月 4 日 —— 周二啦！