[Swift]LeetCode477. 汉明距离总和 | 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.

---

两个整数的 汉明距离 指的是这两个数字的二进制数对应位不一样的数量。

计算一个数组中，任意两个数之间汉明距离的总和。

示例:

输入: 4, 14, 2

输出: 6

解释: 在二进制表示中，4表示为0100，14表示为1110，2表示为0010。（这样表示是为了体现后四位之间关系）
因此答案为：
HammingDistance(4, 14) + HammingDistance(4, 2) + HammingDistance(14, 2) = 2 + 2 + 2 = 6.

注意:

1. 数组中元素的范围为从 0到 10^9。
2. 数组的长度不超过 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 }