Leetcode 338 Counting Bits

本文详细介绍了如何解决计算0到任意整数范围内每个整数二进制形式中1的数量的问题,并通过优化算法提高了效率,避免了使用内置函数,实现了O(n)时间复杂度和O(n)空间复杂度。

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338. Counting Bits

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Total Accepted: 7206  Total Submissions: 13074  Difficulty: Medium

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.

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    Credits:
    Special thanks to @ syedee for adding this problem and creating all test cases.

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      (E) Number of 1 Bits



    这道题是191. Number of 1 Bits的升级版本,需要输出0-N内每个数的二进制1的个数,最终返回vector。


    解法一:O(n * numBits(n))

    这种解法的复杂度是 O(n * numBits(n)) 其中numBits(n)表示n的二进制表示1的位数。

    【函数迭代的次数 = n中二进制1的个数】

    代码如下:

    int bitcount(unsigned int n)
    {
    	int count = 0;
    	while (n)
    	{
    		count++;
    		n &= (n - 1);
    	}
    	return count;
    }

    接下来就是遍历0-n内每个数,调用上面那个函数即可。


    Solution如下(已AC 2016-03-25):

    int bitcount(unsigned int n)
    {
    	int count = 0;
    	while (n)
    	{
    		count++;
    		n &= (n - 1);
    	}
    	return count;
    }
    
    vector<int> countBits(int num) 
    {
    	vector<int> res;
    	res.push_back(0);
    
    	if (num <= 0) return res;
    
    	else
    	{
    		for (int i = 1; i <= num; ++i)
    		{
    			res.push_back(bitcount(i));
    		}
    		return res;
    	}
    }

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