338. Counting Bits

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