338. counting bits

Medium

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 1:

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Input: 2
Output: [0,1,1]

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.

2019.9.12 独立做出来了

方法：

DP + 位运算

对于一个数 n 来说，n&(n-1) 可以把 n 的二进制表示上的最后一个 ‘1’ 删除。

因此 n 的二进制表示中 ‘1’ 的个数 = n&(n-1) 的二进制表示中 ‘1’ 的个数 + 1，由此写 DP。

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class (object):
    def countBits(self, num):
        """
        :type num: int
        :rtype: List[int]
        """
        if num==0:
            return [0]
        ans = [0]*(num+1)
        for i in range(1, num+1):
            temp = i&(i-1)
            if temp==0:
                ans[i] = 1
            else:
                ans[i] = ans[temp]+1
        return ans

类似题目：

Number of 1 Bits