[LeetCode] 338. Counting Bits

350. Intersection of Two Arrays II  

Given a non negative integer number num. For every numbersi in the range0 ≤ 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 timeO(n*sizeof(integer)). But can you do it in linear timeO(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：动规

       话说看到时间空间复杂度要求都为n那基本上就会想到用这个。具体到这道题的话，又有两种思路。

       1. 利用hint，可总结出以下规律：除了0之外，若n为偶数，则其1的个数与n/2相同，因为n的二进制实际上就是n/2的二进制左移一位，而末位为0 。若n为奇数，则1的个数为n/2的个数+1，因为可看做n/2左移一位后加1 。

        

 class Solution { public:    vector<int> countBits(int num) {        vector<int> res;        res.push_back(0);        for(int i=1;i<=num;i++)            if(i%2) res.push_back(res[i/2]+1);            else res.push_back(res[i/2]);        return res;    } };

        2. 利用n&(n-1)。n&(n-1)可以判断n是否为2的整次幂。例如4(100) & 3(011) = 0 。n&(n-1)还可以使n的二进制的末位归零。例如7(111) & 6(110) = 6(110).得到这个后，即可直接利用n&(n-1)的1的个数加1计算结果。

       

class Solution {public:    vector<int> countBits(int num) {        vector<int> res(num + 1, 0);        for (int i = 1; i <= num; ++i) {            res[i] = res[i & (i - 1)] + 1;        }        return res;    }};

        
        PS,如果抛开题目限制，还有很多有意思的计算二进制1的个数的算法。可参考以下文章：

        http://www.cnblogs.com/graphics/archive/2010/06/21/1752421.html