[Swift]LeetCode540. 有序数组中的单一元素 | Single Element in a Sorted Array

Given a sorted array consisting of only integers where every element appears twice except for one element which appears once. Find this single element that appears only once.

Example 1:

Input: [1,1,2,3,3,4,4,8,8]
Output: 2 

Example 2:

Input: [3,3,7,7,10,11,11]
Output: 10 

Note: Your solution should run in O(log n) time and O(1) space.

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给定一个只包含整数的有序数组，每个元素都会出现两次，唯有一个数只会出现一次，找出这个数。

示例 1:

输入: [1,1,2,3,3,4,4,8,8]
输出: 2

示例 2:

输入: [3,3,7,7,10,11,11]
输出: 10

注意: 您的方案应该在 O(log n)时间复杂度和 O(1)空间复杂度中运行。

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Runtime: 40 ms

Memory Usage: 19 MB

 1 class Solution {
 2     func singleNonDuplicate(_ nums: [Int]) -> Int {
 3         var left:Int = 0
 4         var right:Int = nums.count - 1
 5         while (left < right)
 6         {
 7             var mid:Int = left + (right - left) / 2
 8             if mid % 2 == 1 
 9             {
10                 mid -= 1
11             }
12             if nums[mid] == nums[mid + 1]
13             {
14                 left = mid + 2
15             }  
16             else
17             {
18                 right = mid
19             }
20         }
21         return nums[left]
22     }
23 }

---

40ms

 1 class Solution {
 2     func singleNonDuplicate(_ nums: [Int]) -> Int {
 3         var low = 0
 4         var high = nums.count - 1
 5         var medium = (low + high)/2
 6         
 7         while (low < high) {
 8             if medium % 2 == 0 {
 9                 if nums[medium] == nums[medium + 1] {
10                     low = medium + 2
11                 } else {
12                     high = medium
13                 }
14             } else {
15                 if nums[medium] == nums[medium - 1] {
16                     low = medium + 1
17                 } else {
18                     high = medium
19                 }
20             }
21             medium = (low + high)/2
22         }
23         
24         return nums[low]
25     }
26 }

---

Runtime: 40 ms

Memory Usage: 19 MB

 1 class Solution {
 2     func singleNonDuplicate(_ nums: [Int]) -> Int {
 3  var left = 0
 4         var right = nums.count
 5         
 6         var middle = nums.count / 2
 7         while middle != 0  {
 8             guard middle - 1 >= 0 && middle + 1 <= nums.count-1 else
 9             {
10                 return nums[middle]
11             }
12             if nums[middle] == nums[middle-1]{
13                 if middle % 2 == 0{
14                     right = middle - 1
15                 }else{
16                     left = middle + 1
17                 }
18                 
19             }else if nums[middle] == nums[middle+1] {
20                 if middle % 2 == 0{
21                     left = middle + 1
22                 }else
23                 {
24                     right = middle-1
25                 }
26                 
27             }else{
28                 return nums[middle]
29             }
30             middle = (left + right)/2
31         }
32         return nums[0]
33     }
34 }

---

48ms

 1 class Solution {
 2     func singleNonDuplicate(_ nums: [Int]) -> Int {
 3         var dictionary = [Int: Int]()
 4         
 5         for number in nums {
 6             var value = dictionary[number] ?? 0
 7             value += 1
 8             dictionary[number] = value
 9         }
10         
11         var result = 0
12         
13         for key in dictionary.keys where dictionary[key] == 1 {
14             result = key
15             break
16         }
17         
18         return result
19     }
20 }

---

68ms

 1 class Solution {
 2     func singleNonDuplicate(_ numbers: [Int]) -> Int {
 3         func search(_ left: Int, _ right: Int) -> Int {
 4             let middle = (left + right) / 2
 5 
 6             if middle > 0 && numbers[middle - 1] == numbers[middle] {
 7                 if (middle - 1) % 2 == 0 {
 8                     return search(middle + 1, right)
 9                 } else {
10                     return search(left, middle - 2)
11                 }
12             }
13 
14             if middle < (numbers.count - 1) && numbers[middle] == numbers[middle + 1] {
15                 if middle % 2 == 0 {
16                     return search(middle + 2, right)
17                 } else {
18                     return search(left, middle - 1)
19                 }
20             }
21 
22             return numbers[middle]
23         }
24         return search(0, numbers.count - 1)
25     }
26 }