[Swift]LeetCode137. 只出现一次的数字 II | Single Number II

★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★
➤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/9963667.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 non-empty array of integers, every element appears three times except for one, which appears exactly once. Find that single one.

Note:

Your algorithm should have a linear runtime complexity. Could you implement it without using extra memory?

Example 1:

Input: [2,2,3,2]
Output: 3

Example 2:

Input: [0,1,0,1,0,1,99]
Output: 99

---

给定一个非空整数数组，除了某个元素只出现一次以外，其余每个元素均出现了三次。找出那个只出现了一次的元素。

说明：

你的算法应该具有线性时间复杂度。 你可以不使用额外空间来实现吗？

示例 1:

输入: [2,2,3,2]
输出: 3

示例 2:

输入: [0,1,0,1,0,1,99]
输出: 99

---

12ms

class Solution {
    func singleNumber(_ nums: [Int]) -> Int {
        let count = nums.count
        guard count > 0 else { return 0 }
        let range = 1..<count
        var oneTime = nums[0]
        var twoTimes = 0
        var notThreeTimes = 0
        for i in range {
            let next = nums[i]
            twoTimes |= oneTime & next
            oneTime ^= next
            notThreeTimes = ~(oneTime & twoTimes)
            oneTime &= notThreeTimes
            twoTimes &= notThreeTimes
        }
        return oneTime
    }
}

---

16ms

class Solution {
    func singleNumber(_ nums: [Int]) -> Int {
        var a = 0, b = 0;
        for num in nums {
            b = (b ^ num) & ~a;
            a = (a ^ num) & ~b;
        }
        return b;
    }
}

---

24ms

class Solution {
    func singleNumber(_ nums: [Int]) -> Int {
        guard nums.count > 0 else {
            return 0
        }
        
        var temp = [Int: Int]()
        
        for num in nums {
            temp[num] = temp[num, default: 0] + 1
        }
        
        for key in temp.keys {
            if temp[key]! == 1 {
                return key
            } 
        }
        
        return 0
    }
}

---

28ms

class Solution {
    func singleNumber(_ nums: [Int]) -> Int {
        var resSet = Set<Int>(nums)
        var showed = Set<Int>()
        for num in nums {
            if showed.contains(num) {
                resSet.remove(num)
            }
            showed.insert(num)
        }
        return resSet.first!
    }
}

---

76ms

class Solution {
    func singleNumber(_ nums: [Int]) -> Int {
        var numDict = [Int: Int]()
        for num in nums {
            if let value = numDict[num] {
                numDict[num] = value + 1
            } else {
                numDict[num] = 1
            }
        }
            
        let singleVal = numDict.filter { $0.value == 1 }.first!
        return singleVal.key
    }
}

---

80ms

class Solution {
    func singleNumber(_ nums: [Int]) -> Int {
        return ((Set(nums).reduce(0,+) * 3) - nums.reduce(0,+))/2
    }
}

---

80ms

class Solution {
    func singleNumber(_ nums: [Int]) -> Int {
        guard nums.count > 0 else {
            return 0
        }
        var result = 0
        for i in 0..<MemoryLayout<Int>.size * 8 {
            var count = 0
            for j in 0..<nums.count {
                count += (nums[j] >> i) & 1
            }
            
            result += (count % 3) << i
        }
        
        return result
    }
}