Leetcode 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.

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?

class Solution(object):
    def countBits(self, num):
        a = [0 for i in range(num+1)]
        for i in range(1,num+1):
            if i%2 == 0:
                a[i] = a[i/2]
            else:
                a[i] = a[i-1]+1
        return a

Leetcode 338 Counting Bits

原文：http://www.cnblogs.com/lilixu/p/5473275.html