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  • 338. Counting Bits (Medium)

    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.

    思路:
    1.按位与运算(1&1=1, 1&0=0) :res[n] = res[n & (n - 1)] + 1
    2.移位运算:ans[n] = ans[n >> 1] + (n & 1)

    注意:后面加逗号;

    class Solution():
        def countBits(self, num):
            """
            :type num: int
            :rtype: List[int]
            """
            res = [0]
            for i in range(1, num + 1):
                res += res[i & (i - 1)] + 1,
                # res += res[i >>1] + (i & 1),
            return res
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  • 原文地址:https://www.cnblogs.com/yancea/p/7527359.html
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