leetcode：Counting Bits

最新推荐文章于 2019-12-07 14:42:11 发布

Ray_sysu

最新推荐文章于 2019-12-07 14:42:11 发布

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本文链接：https://blog.youkuaiyun.com/Ray_sysu/article/details/74857912

本文介绍了一种高效计算从0到给定整数范围内每个数二进制表示中1的数量的方法。通过分析二进制数的特点，提出了线性时间复杂度O(n)的解决方案。

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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.

给定num，对于每个 0<= i <=num ，计算i的2进制中1的个数。

若对每个数，进行每位的判断，时间复杂度是O(n*logn)。

可以从每个数的最低位开始分析，观察到

100

101

110

111

1000

例如1001001 ，它的二进制1的个数等于100100 总二进制个数 + 1 。即循环地在每个数后面加0、1可得接下来的数字。

因此第i位就是第i/2位+（i%2）的值。

代码：

class Solution {
public:
vector<int> countBits(int num) {
vector<int>vec(num+1,0);
vec[0]=0;
for(int i = 1; i <= num; i++){
vec[i]=vec[i/2]+(i % 2);
}
return vec;
}
};