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.
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update v3.0: class Solution { public: vector<int> countBits(int num) { vector<int> v(num + 1); v[0] = 0; for(int i = 1; i < num + 1; i++) { v[i] = v[i >> 1] + i % 2; } return v; } }; update v2.0: class Solution { public: vector<int> countBits(int num) { vector<int> v(num + 1); v[0] = 0; v[1] = 1; v[2] = 1; v[3] = 2; for(int i = 4; i < num + 1; i = i * 2) { for(int j = i; j < i + i/2 && j < num + 1; j++) { v[j] = v[j - i/2]; } for(int j = i + i/2; j < i * 2 && j < num + 1; j++) { int temp = j; int cnt = 0; while(temp) { cnt = cnt + temp % 2; temp = temp/2; } v[j] = cnt; } } return v; } }; version v1.0: class Solution { public: vector<int> countBits(int num) { vector<int> v(num + 1); for(int i = 0; i <= num; i++) { int temp = i; int cnt = 0; while(temp) { cnt = cnt + temp % 2; temp = temp / 2; } v[i] = cnt; } return v; } }; |
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