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[5/26] 191. Number of 1 Bits

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[LeetCode 시즌 3] 2022년 5월 26일 문제입니다.

https://leetcode.com/problems/number-of-1-bits/


191. Number of 1 Bits
Easy
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Write a function that takes an unsigned integer and returns the number of '1' bits it has (also known as the Hamming weight).

Note:

  • Note that in some languages, such as Java, there is no unsigned integer type. In this case, the input will be given as a signed integer type. It should not affect your implementation, as the integer's internal binary representation is the same, whether it is signed or unsigned.
  • In Java, the compiler represents the signed integers using 2's complement notation. Therefore, in Example 3, the input represents the signed integer. -3.

 

Example 1:

Input: n = 00000000000000000000000000001011
Output: 3
Explanation: The input binary string 00000000000000000000000000001011 has a total of three '1' bits.

Example 2:

Input: n = 00000000000000000000000010000000
Output: 1
Explanation: The input binary string 00000000000000000000000010000000 has a total of one '1' bit.

Example 3:

Input: n = 11111111111111111111111111111101
Output: 31
Explanation: The input binary string 11111111111111111111111111111101 has a total of thirty one '1' bits.

 

Constraints:

  • The input must be a binary string of length 32.

 

Follow up: If this function is called many times, how would you optimize it?

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mingki님의 댓글

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C++
Runtime: 0 ms, faster than 100.00% of C++ online submissions for Number of 1 Bits.
Memory Usage: 6.2 MB, less than 48.74% of C++ online submissions for Number of 1 Bits.
class Solution {
    unordered_map<uint32_t, int> memo;
    
public:
    int hammingWeight(uint32_t n) {
        int cnt = 0;
        if (memo.count(n)) {
            return memo[n];
        }
        memo[n] = 0;
        for (int i = 0; i < 32; ++i) memo[n] += (n >> i) & 1;
        return memo[n];
    }
};
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