[Medium] Counting Bits

【问题】

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.

【解法】

观察二进制数的规律可以知道。对于从0~2^n-1这2^n个数组成的数列a[]，可以知道，前2^(n-1)个数的最低（n-1）位，与后2^(n-1)个数（n-1）位具有相同的发展规律，不同的是前2^(n-1)个数最高位是0，后2^(n-1)个数最高位为1.

由此，可以知道，对于从0~2^n-1这2^n个数组成的数列，取i=2^(n-1),2^(n-1)+1,...,2^n - 1。对于第i个数a[i]，其‘1’的个数是第i-2^(n-1)个数a[i-2^(n-1)]加一。这两个数的区别仅在于最高位，前者为1，后者为0.

【源码】

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