338. Counting Bits

338. Counting Bits

Easy

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Given an integer n, return an array ans of length n + 1 such that for each i (0 <= i <= n), ans[i] is the number of 1's in the binary representation of i.

Example 1:

Input: n = 2
Output: [0,1,1]
Explanation:
0 --> 0
1 --> 1
2 --> 10

Example 2:

Input: n = 5
Output: [0,1,1,2,1,2]
Explanation:
0 --> 0
1 --> 1
2 --> 10
3 --> 11
4 --> 100
5 --> 101

Constraints:

• 0 <= n <= 105

Follow up:

• It is very easy to come up with a solution with a runtime of O(n log n). Can you do it in linear time O(n) and possibly in a single pass?
• Can you do it without using any built-in function (i.e., like __builtin_popcount in C++)?

class Solution:
    def countBits(self, n: int) -> List[int]:
        """
        assert Solution().countBits(5) == [0, 1, 1, 2, 1, 2]
        assert Solution().countBits(2) == [0, 1, 1]
        
        参考解题思路：动态规划和位运算求解，dp[i]代表i二进制有几个1，
        若i二进制位结尾为0，则dp[i]=dp[i>>1]，因为末尾是0了，1的个数与i算术右移一样
        若i二进制位结尾为1，则dp[i]=dp[i-1]+1，因为末尾是1了，就是让i-1的二进制位1的个数加上现在末尾一个1
        时间复杂度：O(n)，空间复杂度：O(n)
        """

        dp = [0] * (n + 1)
        for i in range(1, n + 1):
            dp[i] = dp[i - 1] + 1 if i & 1 else dp[i >> 1]
        return dp