leetcode338

public class Solution {
    public int[] CountBits(int num) {
        var ary = new int[num + 1];

            for (int i = 0; i <= num; i++)
            {
                var count = 0;
                var cur = i;
                do
                {
                    var c = cur % 2;
                    if (c == 1)
                    {
                        count++;
                    }
                    cur = cur / 2;
                } while (cur != 0);

                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 + 1];
            for (int i = 0; i <= num; i++)
            {
                int n = i;
                int count = 0;
                while (n != 0)
                {
                    if ((n & 1) == 1)
                    {
                        count++;
                    }
                    n >>= 1;
                }
                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