191. Number of 1 Bits
一、題目
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?
二、程式作法
/*寫法二 每次取MOD的作用是由最低位向高位一路做二進制的轉換*/
public class Solution
{
public int HammingWeight(uint n)
{
int counter = 0;
while (n > 0)
{
if (n % 2 == 1)
counter++;
n /= 2;
}
return counter;
}
}
/*寫法一 先將輸入轉成二進制表示，然後去數 bit 為 1 的數量*/
/*
public class Solution
{
public int HammingWeight(uint n)
{
string s = Convert.ToString(n, 2);
char[] chars = s.ToCharArray();
int counter = 0;
for (int i = 0; i < chars.Length; i++)
if (chars[i] == '1')
counter++;
return counter;
}
}
*/
三、思路筆記
題目為回傳一無符號整數二進制表示，其中 bit 為 1 的數量。
解法一，先將輸入轉成二進制表示，然後去數 bit 為 1 的數量。
解法二，針對一整數轉換成二進制的過程中，每 1 bit 的轉換都可知道該 bit 為 0 或為 1，進而去計數。