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.
利用或与加中的
a + b = a | b
对每一个数 n，等于比其小的最大的2的指数 a 与 n - a 的和，此时 n = a | (n - a) = a + (n - a)
所以有，ans[n] = ans[a] + ans[n - a]
class Solution {
public:
vector<int> countBits(int num) {
vector<int> ans(num + 1);
if (num == 0) {
return ans;
}
ans[1] = 1;
int bit = 2; // 记录比i大的最小的2的指数
for (size_t i = 2; i <= num; i++) {
if (bit & i) {
bit <<= 1;
ans[i] = 1;
continue;
}
int a = bit >> 1; // a就是比i小的最大的2的指数
int b = i - a;
ans[i] = ans[a] + ans[b];
}
return ans;
}
};
