LeetCode 338. Counting Bits

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题目

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]

Example 2:

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Input: 5
Output: [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

DP.
Or does the odd/even status of the number help you in calculating the number of 1s?
If i is even, then res[i] = res[i/2]
If i is odd, then res[i] = res[i-1] + 1 = res[i/2] + 1
So, res[i] = res[i/2] + i % 2
Aka, res[i] = res[i >> 1] + (i & 1)

代码

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