LeetCode-338.Counting Bits

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.

Hint:

1. You should make use of what you have produced already.
2. Divide the numbers in ranges like [2-3], [4-7], [8-15] and so on. And try to generate new range from previous.
3. Or does the odd/even status of the number help you in calculating the number of 1s?

列举一些就可以找到规律

二进制每多一位就把结果数组中所有的结果都加一

比如n=11(1011)，就是把首位1去掉后结果数组中第3(11)个数加一

方法1

vector<int> countBits(int num)    {        vector<int> res;    res.push_back(0);    if (num == 0)    return res;    for (int i = 1; i <= num; i++)        res.push_back(res[i - (int)pow(2, (int)log2(i))] + 1);    return res;    }

循环体中可以换成更简洁的代码

res.push_back(res[i - (1 << (int)log2(i))] + 1);

方法2

利用n&(n-1) 去掉最低位1，嫌弃去掉最低位1后的1的个数

res.push_back(res[i&(i-1)] + 1);

方法3

先求高位的1个数，再加末位

res.push_back(res[i>>1] + (i&1));