LeetCode_338_CountingBits

338. 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.

题目分析:

对于一个给定的数字num, 给出[0,num]区间内的每个数的二进制表示所含的1的个数，用一个数组输出结果。

尝试给出符合以下条件的solution

1、 T(n) = O(n)

2、 S(n) = O(n)

3、 不使用C++或其他语言的内置函数

解:

假定rst[num] 即为num的二进制表示法中包含的1的个数，存在以下公式

rst[num] = rst[num^(num-1)] + 1

且rst[0] = 0;

Solution:

vector<int> countBits(int num)
{
    vector<int> ret(num + 1, 0);
    for (int i = 1; i <= num; ++i)
        ret[i] = ret[i&(i - 1)] + 1;
    return ret;
}

T(n) = O(n)

S(n) = O(n)

LeetCode_338_CountingBits

原文：http://www.cnblogs.com/HellcNQB/p/5361593.html