338_Counting_Bits

Counting Bits

Difficulty Medium

tags bit mask

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.

思路：
每个数字的一部分都是由比它更小的数组构成的
solution 1

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

trick. fancy.

solution 2

class Solution {
public:
    vector<int> countBits(int num) {
        vector<int> count(1, 0);
        for (int i=1; i < num+1; i++)
            count.push_back(count[i&(i-1)]+1);
        return count;
     }
};