LeetCode 338. Counting Bits

最新推荐文章于 2025-09-10 15:16:07 发布

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原文链接：http://www.cnblogs.com/pegasus923/p/5499737.html

文章标签：

#c/c++

本文针对LeetCode上的计数位问题提供了一种高效的解决方案，采用C++实现，通过观察规律得出每个数的二进制中1的个数等于前一个数的一半的1的个数加上自身奇偶性判断，最终实现线性时间复杂度。

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

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.

vector - C++ Reference
- 　　http://www.cplusplus.com/reference/vector/vector/?kw=vector
二分分治。捡起vector用法。。。

 1 #include <iostream>
 2 #include <vector>
 3 using namespace std;
 4 
 5 class Solution {
 6 public:
 7     vector<int> countBits(int num) 
 8     {
 9                 vector<int> res(num + 1, 0);
10                 
11                 for (int i = 0; i < res.size(); i ++)
12                 {
13                     res[i] = res[i / 2] + i % 2;                    
14                 }        
15                 
16                 return res;
17     }
18 };
19 
20 int main ()
21 {
22     Solution testSolution;
23     vector<int> result(5, 0);
24     
25     result = testSolution.countBits(5);    
26 
27     for (int i = 0; i < result.size(); i ++)
28         cout << result[i] << endl;
29 
30     char ch;
31     cin >> ch;
32     
33     return 0;
34 }