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.

Hint:

1. You should make use of what you have produced already。

因为最近在研究树状数组，被他的精妙所震撼。他巧妙的将一个数的二进制0与1的个数赋予新的含义。

int LowBit(int x){return x & (-x);}

树状数组学习笔记：

http://blog.csdn.net/x_iya/article/details/8943264

看到了这个题，很自然的联想到LowBit的使用。

思路：记忆化搜索+LowBit

对于LowBit(x)：

假设x的末尾0的个数为k则LowBit是计算2^k，同时也是计算末尾1在的位置（即所表示的数）

ibit2^k

101121023111410045101161102711118100089100111010102111011112110041311011141110215111111610000160特殊处理下，直接输出0

当i与2^k相等时，意味着只有一个1，输出1

当i与2^k不相等时，比如我们看i=15，此时2^k=1，即末尾的1代表1，我们把末尾的1“去掉”，1111-->1110

即转化为求14的二进制数1的个数（当然得加上删除的1），也就是我们得利用先前求得的来计算i的二进制数1的个数。

#include <iostream>#include <vector>using namespace std;class Solution {public:int LowBit(int x){return x & (-x);}    vector<int> countBits(int num) {        vector<int>vec;        vec.push_back(0);        for (int i = 1; i <= num; ++i)        {        if (i == LowBit(i))        {        vec.push_back(1);        }        else        {        vec.push_back(vec[i - LowBit(i)] + 1);        }        }        return vec;    }};int main(){Solution s;vector<int>vec = s.countBits(16);for (auto iter = vec.begin(); iter != vec.end(); iter++){cout << *iter << endl;}    return 0;}

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