[Swift]LeetCode448. 找到全部数组中消失的数字 | Find All Numbers Disappeared in an Array

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

Find all the elements of [1, n] inclusive that do not appear in this array.

Could you do it without extra space and in O(n) runtime? You may assume the returned list does not count as extra space.

Example:

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

Output:
[5,6]

---

给定一个范围在  1 ≤ a[i] ≤ n ( n = 数组大小 ) 的 整型数组，数组中的元素一些出现了两次，另外一些只出现一次。

找到全部在 [1, n] 范围之间没有出如今数组中的数字。

您能在不使用额外空间且时间复杂度为O(n)的状况下完成这个任务吗? 你能够假定返回的数组不算在额外空间内。

示例:

输入:
[4,3,2,7,8,2,3,1]

输出:
[5,6]

---

80ms

 1 class Solution {
 2     func findDisappearedNumbers(_ nums: [Int]) -> [Int] {
 3         
 4         var input = nums
 5         var result: [Int] = []
 6         
 7         for i in 0..<input.count{
 8             var check = -1
 9             
10             if input[i] > 0{
11                 check = input[i] - 1
12             }else{
13                 check = (input[i] * (-1)) - 1
14             }
15             
16             if input[check] > 0{
17                 input[check] = (input[check] * (-1))
18             }
19             
20         }
21         
22         for i in 0..<input.count{
23             if input[i] > 0 {
24                 result.append(i+1)
25             }
26         }
27         
28         return result
29     }
30 }

---

88ms

 1 class Solution {
 2     func findDisappearedNumbers(_ nums: [Int]) -> [Int] {
 3         var list = nums
 4         var results = [Int]()
 5 
 6         for i in 0 ..< list.count {
 7             let idx = abs(list[i]) - 1
 8             list[idx] = (list[idx] > 0) ? -list[idx] : list[idx]
 9         }
10 
11         for i in 0 ..< list.count {
12             if (0 < list[i]) {
13                 results.append(i + 1)
14             }
15         }
16 
17         return results
18     }
19 }

---

104ms

 1 class Solution {
 2     func findDisappearedNumbers(_ nums: [Int]) -> [Int] {
 3         let test = 0
 4         if test == 0 {        
 5             var result = [Int]()
 6             var explored = [Bool](repeating: false, count: nums.count+1)
 7 
 8             for num in nums {
 9                 explored[num] = true
10             }
11 
12             var i = 1
13             while i < explored.count {
14                 if explored[i] == false {
15                     result.append(i)
16                 }
17 
18                 i += 1
19             }
20 
21             return result
22         }
23         
24         if test == 1 {        
25             var result = [Int]()
26             var nums = nums
27 
28             var i = 0
29             while i < nums.count {
30                 while nums[i] != nums[nums[i]-1] {
31                     nums.swapAt(i, nums[i]-1)
32                 }
33 
34 
35                 i += 1
36             }
37 
38             var j = 0
39             while j < nums.count {
40                 if j+1 != nums[j] {
41                     result.append(j+1)
42                 }
43 
44                 j += 1
45             }
46 
47             return result
48         }
49     }
50 }

---

112ms

 1 class Solution {
 2     func findDisappearedNumbers(_ nums: [Int]) -> [Int] {
 3         var nums = nums
 4         for idx in 0..<nums.count {
 5             let nextIdx = abs(nums[idx]) - 1
 6             nums[nextIdx] = nums[nextIdx] < 0 ? nums[nextIdx] : -nums[nextIdx]
 7         }
 8         var ans = [Int]()
 9         for idx in 0..<nums.count {
10             if nums[idx] > 0 {
11                 ans.append(idx + 1)
12             }
13         }
14         return ans
15     }
16 }

---

124ms

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