leetcode338

public class Solution {
    public int[] CountBits(int num) {
        var ary = new int[num + ];

            for (int i = ; i <= num; i++)
            {
                var count = ;
                var cur = i;
                do
                {
                    var c = cur % ;
                    if (c == )
                    {
                        count++;
                    }
                    cur = cur / ;
                } while (cur != );

                ary[i] = count;
            }

            return ary;
    }
}

https://leetcode.com/problems/counting-bits/#/description

另一个版本，246ms：

public class Solution
    {
        public int[] CountBits(int num)
        {
            int[] counts = new int[num + ];
            for (int i = ; i <= num; i++)
            {
                int n = i;
                int count = ;
                while (n != )
                {
                    if ((n & ) == )
                    {
                        count++;
                    }
                    n >>= ;
                }
                counts[i] = count;
            }
            return counts;
        }
    }

补充一个使用动态规划思想的代码，使用python实现：

 class Solution:
     def countBits(self, num: 'int') -> 'List[int]':
         if num==0:
             return [0]
         elif num==1:
             return [0,1]
         else:
             d = [0] * (num+1)
             d[0] = 0
             d[1] = 1
             for i in range(2,num+1):
                 if i % 2 ==0:
                     d[i] = d[i//2]
                 else:
                     d[i] = d[i//2] + 1
             return d