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 in c++ or in any other language.

Hint:

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

Solution

一个数 * 2 就是把它的二进制全部左移一位，也就是说 1的个数是相等的。

那么我们可以利用这个结论来做。

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class Solution(object):
    def countBits(self, num):
        """
        :type num: int
        :rtype: List[int]
        """
        ret = [0]*(num+1)
        for i in range(1,num+1):
            ret[i] = ret[i>>1]+(i&1)
        return ret

res[i /2] 然后看看最低位是否为1即可（上面*2一定是偶数，这边比如15和14除以2都是7，但是15时通过7左移一位并且+1得到，14则是直接左移）