[Swift]LeetCode442. 數組中重複的數據 | Find All Duplicates in an Array

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Given an array of integers, 1 ≤ a[i] ≤ n (n = size of array), some elements appear twice and others appear once.

Find all the elements that appear twice in this array.

Could you do it without extra space and in O(n) runtime? 

Example:

Input:
[4,3,2,7,8,2,3,1]

Output:
[2,3]

---

給定一個整數數組 a，其中1 ≤ a[i] ≤ n （n爲數組長度）, 其中有些元素出現兩次而其餘元素出現一次。

找到全部出現兩次的元素。

你能夠不用到任何額外空間並在O(n)時間複雜度內解決這個問題嗎？

示例：

輸入:
[4,3,2,7,8,2,3,1]

輸出:
[2,3]

---

136ms

 1 class Solution {
 2     func findDuplicates(_ nums: [Int]) -> [Int] {
 3         var nums = nums
 4         var result = [Int]()
 5         for num in nums {
 6             nums[num - 1] = -nums[num - 1]
 7             if (nums[num - 1] > 0) {
 8                 result.append(num)
 9             }
10         }
11         return result
12     }
13 }

---

144ms

 1 class Solution {
 2     func findDuplicates(_ nums: [Int]) -> [Int] {
 3         var res = [Int]()
 4         var nums = nums
 5         for i in 0..<nums.count {
 6             let index = abs(nums[i]) - 1
 7             if nums[index] < 0 {
 8                 res.append(abs(index + 1))
 9             }
10             nums[index] = -nums[index]
11         }
12         return res
13     }
14 }

---

512ms

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

---

708ms

 1 class Solution {
 2     func findDuplicates(_ nums: [Int]) -> [Int] {
 3         guard nums.count > 1 else {
 4             return []
 5         }
 6         var set = Set<Int>()
 7         var arr = [Int]()
 8         for i in nums {
 9             if set.contains(i) {
10                 arr.append(i)
11             } else {
12                 set.insert(i)
13             }
14         }
15         return arr
16     }
17 }