[leetcode] 338. Counting Bits

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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.  

Solution:

动态规划，如果i是奇数，f[i] = f[i/2] + 1, 偶数, f[i] = f[i/2]

 1 vector<int> countBits(int num) 
 2     {
 3         vector<int> table(num + 1, 0);
 4         
 5         for (int i = 1; i <= num; i++)
 6             if (i & 1)
 7                 table[i] = table[i >> 1] + 1;
 8             else
 9                 table[i] = table[i >> 1]; 
10         
11         return table;
12     }

[leetcode] 338. Counting Bits

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原文地址：http://www.cnblogs.com/ym65536/p/5601880.html