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 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.

 

Approach #1: Math. [C++]

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

　　

Analysis:

i     binary    　　count   　　i&(i-1)

0　　0　　　      0　　

1　　0001　　　1　　　　  0000  ->  0

2　　0010　　　1　　　　  0000　　0

3　　0011　　　2　　　　  0010　　2

4　　0100　　　1　　　　  0000　　0　　

5　　0101　　    2　　　　  0100　　4

6　　0110　　　2　　　　  0100　　4

7　　0111　　　3　　　　  0110　　 6

8　　1000　　　1　　　　  0000　　0

 

We can find some regular about:

the count of '1' in the i(binary) = the count of '1' in the i&(i-1)(binary) + 1;

temp[i] = temp[i&(i-1)] + 1;

 

Approach #2: bitset. [C++]

class Solution {
public:
    vector<int> countBits(int num) {
        vector<int> res;
        for (int i = 0; i <= num; ++i) {
            res.push_back(bitset<32>(i).count());
        }
        return res;
    }
};

　　

Analysis:

There is a trick of the function bitset. Using this function we can directly calculate the number of '1' in the i's binary form.

 

Reference:

http://www.cplusplus.com/reference/bitset/bitset/count/

http://www.cnblogs.com/grandyang/p/5294255.html