[Swift]LeetCode191. 位1的個數 | Number of 1 Bits

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➤微信公衆號:山青詠芝(shanqingyongzhi)
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Write a function that takes an unsigned integer and return the number of '1' bits it has (also known as the Hamming weight).git

Example 1:github

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

Example 2:算法

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

Example 3:微信

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

Note:函數

  • Note that in some languages such as Java, there is no unsigned integer type. In this case, the input will be given as signed integer type and should not affect your implementation, as the internal binary representation of the integer 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 above the input represents the signed integer -3.

Follow up:優化

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


編寫一個函數,輸入是一個無符號整數,返回其二進制表達式中數字位數爲 ‘1’ 的個數(也被稱爲漢明重量)。spa

示例 1:code

輸入:00000000000000000000000000001011
輸出:3
解釋:輸入的二進制串 
00000000000000000000000000001011 中,共有三位爲 '1'。

示例 2:

輸入:00000000000000000000000010000000
輸出:1
解釋:輸入的二進制串 00000000000000000000000010000000 中,共有一位爲 '1'。

示例 3:

輸入:11111111111111111111111111111101
輸出:31
解釋:輸入的二進制串 11111111111111111111111111111101 中,共有 31 位爲 '1'。

提示:

  • 請注意,在某些語言(如 Java)中,沒有無符號整數類型。在這種狀況下,輸入和輸出都將被指定爲有符號整數類型,而且不該影響您的實現,由於不管整數是有符號的仍是無符號的,其內部的二進制表示形式都是相同的。
  • 在 Java 中,編譯器使用二進制補碼記法來表示有符號整數。所以,在上面的 示例 3 中,輸入表示有符號整數 -3

進階:
若是屢次調用這個函數,你將如何優化你的算法?


12ms

 1 class Solution {
 2     func hammingWeight(_ n: UInt32) -> UInt32 {
 3         var n = n
 4         var res: UInt32 = 0
 5         for i in 0..<32
 6         {
 7             res += (n & 1)
 8             n = n >> 1
 9         }
10         return res
11     }
12 }
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