[Swift]LeetCode260. 只出現一次的數字 III | Single Number III

★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★
➤微信公衆號：山青詠芝（shanqingyongzhi）
➤博客園地址：山青詠芝（https://www.cnblogs.com/strengthen/）
➤GitHub地址：https://github.com/strengthen/LeetCode
➤原文地址：http://www.javashuo.com/article/p-ppotppvo-ma.html 
➤若是連接不是山青詠芝的博客園地址，則多是爬取做者的文章。
➤原文已修改更新！強烈建議點擊原文地址閱讀！支持做者！支持原創！
★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★

Given an array of numbers nums, in which exactly two elements appear only once and all the other elements appear exactly twice. Find the two elements that appear only once.

Example:

Input:  
Output: [1,2,1,3,2,5][3,5]

Note:

1. The order of the result is not important. So in the above example, [5, 3] is also correct.
2. Your algorithm should run in linear runtime complexity. Could you implement it using only constant space complexity?

---

給定一個整數數組 nums，其中剛好有兩個元素只出現一次，其他全部元素均出現兩次。 找出只出現一次的那兩個元素。

示例 :

輸入: 
輸出: [1,2,1,3,2,5][3,5]

注意：

1. 結果輸出的順序並不重要，對於上面的例子， [5, 3] 也是正確答案。
2. 你的算法應該具備線性時間複雜度。你可否僅使用常數空間複雜度來實現？

---

76ms

 1 class Solution {
 2     func singleNumber(_ nums: [Int]) -> [Int] {
 3 
 4         var res = [0, 0]
 5         var diff = nums.reduce(0){$0 ^ $1}
 6         diff &= -diff
 7         for num in nums {
 8             if num & diff != 0 {
 9                 res[0] ^= num
10             }
11             else {
12                 res[1] ^= num
13             }
14         }
15         return res
16     }
17 }

---

76ms

 1 class Solution {
 2     func singleNumber(_ nums: [Int]) -> [Int] {
 3         guard nums.count > 1 else {
 4             return nums
 5         }
 6         
 7         var result = nums[0]
 8         for num in nums[1...] {
 9             result ^= num
10         }
11         
12         var num1 = 0, num2 = 0, tmp = 1
13         while (result & tmp) == 0 {
14             tmp <<= 1
15         }
16         
17         for num in nums {
18             if (num & tmp) > 0 {
19                 num1 ^= num
20             }
21             else {
22                 num2 ^= num
23             }
24         }
25         
26         return [num1, num2]
27     }
28 }

---

80ms

 1 class Solution {
 2     func singleNumber(_ nums: [Int]) -> [Int] {
 3         var s: Set<Int> = Set(minimumCapacity: nums.count)
 4         for i in nums {
 5             if s.contains(i) {
 6                 s.remove(i)
 7             } else {
 8                 s.insert(i)
 9             }
10         }
11         return Array(s)
12     }
13 }

---

84ms

 1 class Solution {
 2     func singleNumber(_ nums: [Int]) -> [Int] {
 3         var array = nums.sorted()
 4         // print(array)
 5         var output:[Int] = []
 6         
 7         var i = 0
 8         while i < array.count {
 9             if i == array.count - 1 || array[i] != array[i+1] {
10                 output.append(array[i])
11             } else {
12                 i += 1
13             }
14             i += 1
15             if output.count > 1 {
16                 break
17             }
18         }
19         
20         return output
21     }
22 }

---

84ms

 1 class Solution {
 2     func singleNumber(_ nums: [Int]) -> [Int] {
 3         var result = [Int]()
 4         var temp = nums.sorted()
 5         for i in 0..<temp.count {
 6             if i == 0 {
 7                 if temp[i] != temp[i+1] {
 8                     result.append(temp[i])
 9                 }
10             } else if i == temp.count - 1{
11                 if temp[i] != temp[i-1] {
12                     result.append(temp[i])
13                 }
14             }
15             else {
16                 if temp[i] != temp[i-1] && temp[i] != temp[i+1]{
17                     result.append(temp[i])
18                 }
19             }
20         }
21         return result
22     }
23 }