[Swift]LeetCode961. 重复 N 次的元素 | N-Repeated Element in Size 2N 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/10165080.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！
★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★

In a array A of size 2N, there are N+1 unique elements, and exactly one of these elements is repeated N times.

Return the element repeated N times.

Example 1:

Input: [1,2,3,3]
Output: 3

Example 2:

Input: [2,1,2,5,3,2]
Output: 2

Example 3:

Input: [5,1,5,2,5,3,5,4]
Output: 5

Note:

1. 4 <= A.length <= 10000
2. 0 <= A[i] < 10000
3. A.length is even

---

在大小为 2N 的数组 A 中有 N+1 个不同的元素，其中有一个元素重复了 N 次。

返回重复了 N 次的那个元素。

示例 1：

输入：[1,2,3,3]
输出：3

示例 2：

输入：[2,1,2,5,3,2]
输出：2

示例 3：

输入：[5,1,5,2,5,3,5,4]
输出：5

提示：

1. 4 <= A.length <= 10000
2. 0 <= A[i] < 10000
3. A.length 为偶数

---

276ms

class BitVector {
  private let wordSize: Int = 64
  var vector: [Int]
  
  init(size: Int) {
    let count = abs(size/wordSize)+1
    vector = Array(repeating: 0, count: count)
  }
  
  func setBit(int: Int) -> Bool {
    let wordIndex = abs(int/wordSize)
    let word = vector[wordIndex]
    let mask = 1 << (int%wordSize)
      
    if word & mask != 0 {
      return false
    } else {
      vector[wordIndex] = word | mask
    }
    return true
  }
}

class Solution {
    func repeatedNTimes(_ A: [Int]) -> Int {
        let bitVector = BitVector(size: 10000)
        for a in A {
            if !bitVector.setBit(int: a) {
                return a
            }
        }
        return -1
    }
}

---

280ms

class Solution {
    func repeatedNTimes(_ A: [Int]) -> Int {
        var dict : [Int: Int] = [:]
        for i in A {
            dict[i] = (dict[i] ?? 0) + 1
            if dict[i] == 2 {
                return i
            }
        }
        return -1
    }
}

---

280ms

class Solution {
    func repeatedNTimes(_ A: [Int]) -> Int {
        var set:Set<Int> = Set<Int>()
        for x in A
        {
            if set.contains(x)
            {
                return x
            }
            set.insert(x)
        }
        return -1
    }
}

---

300ms 

class Solution {
    func repeatedNTimes(_ A: [Int]) -> Int {
        var n:Int = A.count
        var f:[Int] = [Int](repeating:0,count:100000)
        for v in A
        {
            f[v] += 1
        }
        
        for i in 0..<100000
        {
            if f[i] > 1
            {
                return i
            }
        }
        return -1
    }
}

---

316ms

class Solution {
    func repeatedNTimes(_ A: [Int]) -> Int {
        let aDict = Dictionary( A.map{ ($0, 1) }, uniquingKeysWith: +)
        for (key, value) in aDict {
            if value >= 2 {
                return key
            }
        }
        return -1
    }
}

---

336ms

class Solution {
    func repeatedNTimes(_ A: [Int]) -> Int {
        var counts = [Int : Int]()
        
        A.forEach( { counts[$0, default: 0] += 1 })
        
        return counts.max(by: {a, b in a.value < b.value })!.key
    }
}

---

396ms

class Solution {
    func repeatedNTimes(_ A: [Int]) -> Int {
       
        var numSet : Set<Int> = []
        
        for num in A.sorted()[0...A.count/2+1]
        {
            if numSet.contains(num)
            {
                return num
            }
            numSet.insert(num)
        }
        return 0
    }
}