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.
2. Divide the numbers in ranges like [2-3], [4-7], [8-15] and so on. And try to generate new range from previous.
3. Or does the odd/even status of the number help you in calculating the number of 1s?

这道题是统计二进制数字num以内每个数中‘1’的个数，题目难度为Medium。

我们先拿num=3为例，结果为[0, 1, 1, 2]，得到3以内数字的结果之后，4~7的结果可以在此基础上确定，如下0~3二进制表示为：

 bin                ret

000                0

001                1

010                1

011                2

4~7二进制表示为：

  bin                ret

100                1

101                2

110                2

111                3

从上面分析可以看出，4~7除了最高位其他位和0~3是相同的，而最高位是1，这样4~7的结果就可以在0~3的基础上分别加1获得，题目就很容易解决了，具体代码：

class Solution {
public:
    vector<int> countBits(int num) {
        vector<int> ret(1, 0);
        int cnt = 0;
        while(cnt < num) {
            int sz = ret.size();
            for(int i=0; i<sz&&cnt<num; ++i,++cnt) {
                ret.push_back(ret[i]+1);
            }
        }
        return ret;
    }
};

 

上面从高位入手找出解决办法，还可以从低位入手。‘1’的个数等于除了最低位之外的‘1’的个数加上最低位‘1’的个数，即ret[n] = ret[n>>1] + n%2，具体代码：

class Solution {
public:
    vector<int> countBits(int num) {
        vector<int> ret(num+1, 0);
        for(int i=1; i<=num; ++i)
            ret[i] = ret[i>>1] + i%2;
        return ret;
    }
};