[Swift]LeetCode477. 漢明距離總和 | Total Hamming Distance

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The Hamming distance between two integers is the number of positions at which the corresponding bits are different.

Now your job is to find the total Hamming distance between all pairs of the given numbers.

Example:

Input: 4, 14, 2

Output: 6

Explanation: In binary representation, the 4 is 0100, 14 is 1110, and 2 is 0010 (just
showing the four bits relevant in this case). So the answer will be:
HammingDistance(4, 14) + HammingDistance(4, 2) + HammingDistance(14, 2) = 2 + 2 + 2 = 6.

 

Note:

1. Elements of the given array are in the range of 0 to 10^9
2. Length of the array will not exceed 10^4.

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兩個整數的 漢明距離 指的是這兩個數字的二進制數對應位不一樣的數量。

計算一個數組中，任意兩個數之間漢明距離的總和。

示例:

輸入: 4, 14, 2

輸出: 6

解釋: 在二進制表示中，4表示爲0100，14表示爲1110，2表示爲0010。（這樣表示是爲了體現後四位之間關係）
因此答案爲：
HammingDistance(4, 14) + HammingDistance(4, 2) + HammingDistance(14, 2) = 2 + 2 + 2 = 6.

注意:

1. 數組中元素的範圍爲從 0到 10^9。
2. 數組的長度不超過 10^4。

---

Runtime: 668 ms

Memory Usage: 8.1 MB

 1 class Solution {
 2     func totalHammingDistance(_ nums: [Int]) -> Int {
 3         var res:Int = 0
 4         var n:Int = nums.count
 5         for i in 0..<32
 6         {
 7             var cnt:Int = 0
 8             for num in nums
 9             {
10                 if num & (1 << i) != 0
11                 {
12                     cnt += 1
13                 }
14             }
15             res += cnt * (n - cnt)
16         }
17         return res
18     }
19 }