Leetcode: 338. Bit Count

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.

 

Credits:
Special thanks to @ syedee for adding this problem and creating all test cases.

Given a non-negative integer num. For each number i in the range 0 ≤ i ≤ num , count the number of 1s in its binary numbers and return them as an array.

Example:
such as given  num = 5 , should return  [0,1,1,2,1,2].

Advanced:

• It is very easy to give a solution with time complexity O(n * sizeof(integer)). But can you do it with one traversal in linear time O(n)?
• The space complexity of the algorithm is required to be O(n).
• Can you refine the solution further? Do not use any built-in functions (like __builtin_popcount in c++) in c++ or any other language to do this.

Acknowledgments:
Special thanks to  @syedee  for adding this question and all the test cases.

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　　In this question, first of all, according to the requirements of the title, we do not consider __builtin_popcount in C++ for direct calculation.

　　To calculate the number of 1's in its binary number, we first write out some non-negative integers and convert them into binary form to discover the pattern.

0   0000　　0
1   0001　　1
2   0010　　1
3   0011　　2
4   0100　　1
5   0101　　2
6   0110　　2
7   0111　　3
8   1000　　1
9   1001　　2
10  1010　　2
11  1011　　3
12  1100　　2
13  1101　　3
14  1110　　3
15  1111　　4

　　According to the written data, we can see two more obvious laws, which leads to two easier methods

　　Method 1: We can first see that each i value is the value corresponding to i&(i-1) plus 1, for example: the number of 4 is the number of values ​​of 4&3 plus 1. Regarding the AND (&) operation, You can find relevant information online, the code is as follows:

 1 class Solution {
 2 public:
 3     vector<int> countBits(int num) {
 4         vector<int> res;
 5         for(int i=1;i<=num;++i)
 6         {
 7             res[i]=res[i&(i-1)]+1;
 8         }
 9         return res;
10     }
11 };

　　Method 2: We can see that the number of 1s in an odd number is the number of 1s in the number obtained by dividing the number by 2 plus 1. For example: the number of 1s corresponding to 3 is the number of 3/2+1 That is, 1+1=2, the code is as follows:

class Solution {
public:
    vector<int> countBits(int num) {
        vector<int> res{0};
        for (int i = 1; i <= num; ++i) {
            if (i % 2 == 0) res.push_back(res[i / 2]);
            else res.push_back(res[i / 2] + 1);
        }
        return res;
    }
};