LeetCode 第 338 题 (Counting Bits)

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.

借助之前算过的位数来计算，假如ret中保存结果，i的二进制有n位，则ret[i]表示i中bit为1的位数，ret[i >> 1]表示去掉i中最低位的结果，ret[i>>1] + (i&1)即为i的前n-1位为1的数量加上自己最后一位是否为1

class Solution {

public:

    vector<int> countBits(int num) {

        vector<int> ret;

        ret.push_back(0);

        for(int i = 1; i <= num; i++)

        {

            int bits = ret[i >> 1] + (i & 1);

            ret.push_back(bits);

        }

        return ret;

    }

};

巴特西

LeetCode 第 338 题 (Counting Bits)

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