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

★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★
➤微信公众号：山青咏芝（shanqingyongzhi）
➤博客园地址：山青咏芝（https://www.cnblogs.com/strengthen/）
➤GitHub地址：https://github.com/strengthen/LeetCode
➤原文地址：https://www.cnblogs.com/strengthen/p/10409012.html 
➤如果链接不是山青咏芝的博客园地址，则可能是爬取作者的文章。
➤原文已修改更新！强烈建议点击原文地址阅读！支持作者！支持原创！
★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★

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.

---

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。

---

Runtime: 40 ms

Memory Usage: 19 MB

class Solution {
    func singleNonDuplicate(_ nums: [Int]) -> Int {
        var left:Int = 0
        var right:Int = nums.count - 1
        while (left < right)
        {
            var mid:Int = left + (right - left) / 2
            if mid % 2 == 1 
            {
                mid -= 1
            }
            if nums[mid] == nums[mid + 1]
            {
                left = mid + 2
            }  
            else
            {
                right = mid
            }
        }
        return nums[left]
    }
}

---

40ms

class Solution {
    func singleNonDuplicate(_ nums: [Int]) -> Int {
        var low = 0
        var high = nums.count - 1
        var medium = (low + high)/2
        
        while (low < high) {
            if medium % 2 == 0 {
                if nums[medium] == nums[medium + 1] {
                    low = medium + 2
                } else {
                    high = medium
                }
            } else {
                if nums[medium] == nums[medium - 1] {
                    low = medium + 1
                } else {
                    high = medium
                }
            }
            medium = (low + high)/2
        }
        
        return nums[low]
    }
}

---

Runtime: 40 ms

Memory Usage: 19 MB

class Solution {
    func singleNonDuplicate(_ nums: [Int]) -> Int {
 var left = 0
        var right = nums.count
        
        var middle = nums.count / 2
        while middle != 0  {
            guard middle - 1 >= 0 && middle + 1 <= nums.count-1 else
            {
                return nums[middle]
            }
            if nums[middle] == nums[middle-1]{
                if middle % 2 == 0{
                    right = middle - 1
                }else{
                    left = middle + 1
                }
                
            }else if nums[middle] == nums[middle+1] {
                if middle % 2 == 0{
                    left = middle + 1
                }else
                {
                    right = middle-1
                }
                
            }else{
                return nums[middle]
            }
            middle = (left + right)/2
        }
        return nums[0]
    }
}

---

48ms

class Solution {
    func singleNonDuplicate(_ nums: [Int]) -> Int {
        var dictionary = [Int: Int]()
        
        for number in nums {
            var value = dictionary[number] ?? 0
            value += 1
            dictionary[number] = value
        }
        
        var result = 0
        
        for key in dictionary.keys where dictionary[key] == 1 {
            result = key
            break
        }
        
        return result
    }
}

---

68ms

class Solution {
    func singleNonDuplicate(_ numbers: [Int]) -> Int {
        func search(_ left: Int, _ right: Int) -> Int {
            let middle = (left + right) / 2

            if middle > 0 && numbers[middle - 1] == numbers[middle] {
                if (middle - 1) % 2 == 0 {
                    return search(middle + 1, right)
                } else {
                    return search(left, middle - 2)
                }
            }

            if middle < (numbers.count - 1) && numbers[middle] == numbers[middle + 1] {
                if middle % 2 == 0 {
                    return search(middle + 2, right)
                } else {
                    return search(left, middle - 1)
                }
            }

            return numbers[middle]
        }
        return search(0, numbers.count - 1)
    }
}