Leetcode Counting Bits 338
2016-12-27 20:38
302 查看
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.
题目链接
不让一个一个的计算
不允许使用已经有的库函数
被逼着查看之间的关系
每次添加的一位不是0就是1
vc[i] = vc[i/2] + (i%2)
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.
题目链接
不让一个一个的计算
不允许使用已经有的库函数
被逼着查看之间的关系
每次添加的一位不是0就是1
vc[i] = vc[i/2] + (i%2)
class Solution { public: vector<int> countBits(int num) { vector<int> vc(num+1,0); for(int i=1;i<=num;i++){ vc[i]=vc[i/2]+(i%2); } return vc; } };
相关文章推荐
- Counting Bits leetcode 338
- leetcode 338 Counting Bits C++
- leetcode 338 c++. Counting Bits
- leetcode338 Counting Bits Java
- LeetCode Counting Bits
- leetcode oj java Counting Bits
- LeetCode Counting Bits
- LeetCode "Counting Bits"
- LeetCode Counting Bits
- leetcode Counting Bits
- leetcode:bits:Counting Bits(338)
- leetcode Counting Bits
- leetcode之counting bits改进算法
- Leetcode Counting Bits
- leetcode: (190) Reverse Bits
- leetcode 338. Counting Bits
- Number of 1 Bits--LeetCode
- LeetCode 190. Reverse Bits
- Leetcode算法学习日志-338 Counting Bits
- leetcode 338. Counting Bits