[Swift]LeetCode442. 数组中重复的数据 | 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]

---

给定一个整数数组 a，其中1 ≤ a[i] ≤ n （n为数组长度）, 其中有些元素出现两次而其余元素出现一次。

找到全部出现两次的元素。

你能够不用到任何额外空间并在O(n)时间复杂度内解决这个问题吗？

示例：

输入:
[4,3,2,7,8,2,3,1]

输出:
[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 }