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:
- You should make use of what you have produced already.
- Divide the numbers in ranges like [2-3], [4-7], [8-15] and so on. And try to generate new range from previous.
- Or does the odd/even status of the number help you in calculating the number of 1s?
1、Solution 1
class Solution {
public:
vector<int> countBits(int num) {
vector<int> result;
for(int i=0;i<=num;i++){
int count=bitnum(i);
result.push_back(count);
}
return result;
}
int bitnum(int n){
int num=0;
while(n>0){
num+=(n&1);
n>>=1;
}
return num;
}
};
2 、solution 2
class Solution {
public:
vector<int> countBits(int num) {
vector<int> result(num+1);
for(int i=0;i<=num;i++){
if(i%2==0){
result[i]=result[i/2];
}else{
result[i]=result[i/2]+1;
}
}
return result;
}
};