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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)

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;
}
};
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标签:  leetcode