leetcode：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.

给定num，对于每个 0<= i <=num ， 计算i的2进制中1的个数。

若对每个数，进行每位的判断，时间复杂度是O(n*logn)。

可以从每个数的最低位开始分析，观察到

1 10 11 100 101 110 111 1000 例如1001001 ，它的二进制1的个数等于100100 总二进制个数 + 1 。即循环地在每个数后面加0、1可得接下来的数字。 因此第i位就是第i/2位+（i%2）的值。 代码： class Solution {   public:       vector<int> countBits(int num) {           vector<int>vec(num+1,0);           vec[0]=0;           for(int i = 1; i <= num; i++){                   vec[i]=vec[i/2]+(i % 2);               }           return vec;       }   };