Problem:
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.
思路：
Solution (C++):
vector<int> countBits(int num) {
vector<int> res(num+1, 0);
for (int i = 0; i <= num; ++i) {
res[i] = get_bit(i);
}
return res;
}
int get_bit(int n) {
if (n == 0) return 0;
int count = 0;
while (n) {
if (n%2 == 1) count++;
n /= 2;
}
return count;
}
性能：
Runtime: 160 ms Memory Usage: 7.1 MB
思路：
Solution (C++):
vector<int> countBits(int num) {
vector<int> res(num+1, 0);
for (int i = 1; i <= num; ++i) {
res[i] = res[i&(i-1)] + 1;
}
return res;
}
性能：
Runtime: 72 ms Memory Usage: 7.2 MB
