[Swift]LeetCode645. 错误的集合 | Set Mismatch

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The set S originally contains numbers from 1 to n. But unfortunately, due to the data error, one of the numbers in the set got duplicated to another number in the set, which results in repetition of one number and loss of another number.

Given an array nums representing the data status of this set after the error. Your task is to firstly find the number occurs twice and then find the number that is missing. Return them in the form of an array.

Example 1:

Input: nums = [1,2,2,4]
Output: [2,3] 

Note:

1. The given array size will in the range [2, 10000].
2. The given array's numbers won't have any order.

---

集合 S 包含从1到 n 的整数。不幸的是，因为数据错误，导致集合里面某一个元素复制了成了集合里面的另外一个元素的值，导致集合丢失了一个整数并且有一个元素重复。

给定一个数组 nums 代表了集合 S 发生错误后的结果。你的任务是首先寻找到重复出现的整数，再找到丢失的整数，将它们以数组的形式返回。

示例 1:

输入: nums = [1,2,2,4]
输出: [2,3]

注意:

1. 给定数组的长度范围是 [2, 10000]。
2. 给定的数组是无序的。

---

Runtime: 204 ms

Memory Usage: 19.5 MB

class Solution {
    func findErrorNums(_ nums: [Int]) -> [Int] {
        var nums = nums
        for i in 0..<nums.count
        {
            while(nums[i] != nums[nums[i] - 1])
            {
                nums.swapAt(i,nums[i] - 1)
            }
        }
        for i in 0..<nums.count
        {
            if nums[i] != i + 1
            {
               return [nums[i], i + 1]
            }
        }
        return []
    }
}

---

252ms

class Solution {
    func findErrorNums(_ nums: [Int]) -> [Int] {
        var countArr = [Int](repeating: 0,count: nums.count )
       
        for i in 0..<nums.count{
            countArr[i] = -1
        }
        for i in 0..<nums.count{
            countArr[nums[i] - 1] = countArr[nums[i] - 1] + 1
        }
        var absent=0
        var duplicate=0
        
        for i in 0..<nums.count{
            if(countArr[i] == -1){
                absent = i
            }
            if(countArr[i] == 1){
                duplicate = i
            }
        }
        return [duplicate + 1, absent+1]
    }
}

---

256ms

class Solution {
  func findErrorNums(_ nums: [Int]) -> [Int] {
    var nums = nums
    
    var errorNums: [Int] = []
    
    for num in nums {
      let n = abs(num)
      if nums[n - 1] < 0 {
        errorNums.append(n)
      } else {
        nums[n - 1] *= -1
      }
    }
    
    for i in 0..<nums.count {
      if nums[i] > 0 {
        errorNums.append(i + 1)
        break
      }
    }
    
    return errorNums
  }
}

---

260ms

class Solution {
    func findErrorNums(_ nums: [Int]) -> [Int] {
        var counter = Array(repeating: 0, count: nums.count)
        
        var dup = 0
        for n in nums {
            if counter[n - 1] == 1 {
                dup = n
                break
            }
            counter[n - 1] = 1
        }
        
        let l = nums.count
        var missing = (1 + l) * l / 2 - nums.reduce(0, +) + dup
        return [dup, missing]
    }
}

---

272ms

class Solution {
    func findErrorNums(_ nums: [Int]) -> [Int] {
        var array = nums
        var dup = 0
        var missing = 0
        for i in 0..<nums.count {
            let index = abs(nums[i]) - 1
            if array[index] < 0 {
                dup = index + 1
            } else {
                array[index] *= -1
            }
        }
        for i in 0..<array.count {
            if array[i] > 0 {
                missing = i + 1
                break
            }
        }
        return [dup, missing]
    }
}

---

280ms

class Solution {
    func findErrorNums(_ nums: [Int]) -> [Int] {
        var set: Set<Int> = []
        var dup = 0
        let n = nums.count
        var compSum = 0
        for num in nums {
            if dup == 0 && set.contains(num) {
                dup = num
            }
            if dup == 0 {
                set.insert(num)
            }
            compSum += num
        }
        let sum = (n * (n + 1))/2
        let diff = sum - compSum
        return [dup, dup + diff]
    }
}

---

296ms

class Solution {
    func findErrorNums(_ nums: [Int]) -> [Int] {
        var set = Set<Int>()
        
        var result = [Int]()
        
        for n in nums {
            if set.contains(n) {
                result.append(n)
            }
            set.insert(n)
        }
        
        for i in 1...nums.count {
            if !set.contains(i) {
                result.append(i)
            }
        }
        return result
    }
}

---

324ms

class Solution {
  func findErrorNums(_ nums: [Int]) -> [Int] {
    var nums = nums
    
    var errorNums: [Int] = []
    
    for i in 0..<nums.count {
      if nums[abs(nums[i]) - 1] < 0 {
        errorNums.append(abs(nums[i]))
      } else {
        nums[abs(nums[i]) - 1] *= -1
      }
    }
    
    for i in 0..<nums.count {
      if nums[i] > 0 {
        errorNums.append(i + 1)
        break
      }
    }
    
    return errorNums
  }
}

---

340ms

class Solution {
    func findErrorNums(_ nums: [Int]) -> [Int] {
        var set = Set(1...nums.count)
        var dup = -1
        for x in nums {
            if !set.contains(x) {
                dup = x
            }
            set.remove(x)
        }
        return [dup, set.first!]
    }
}

---

380ms

class Solution {
    func findErrorNums(_ nums: [Int]) -> [Int] {
        var seen = Set<Int>();
        
        var dup : Int = 0
        var sm = 0
        for n in nums {
            if seen.contains(n) {
                dup = n
            }
            seen.insert(n)
            sm += n
        }
        let t = nums.count*(nums.count+1)/2-sm+dup
        return [dup, t]
    }
}

---

420ms

class Solution {
    func findErrorNums(_ nums: [Int]) -> [Int] {
        var sum = Array(1...nums.count).reduce(0, { $0 + $1 })
        var set = Set<Int>()
        var res = [Int]()

        for num in nums {
            if set.contains(num) {
                res.append(num)
            } else {
                set.insert(num)
                sum -= num
            }
        }
        res.append(sum)

        return res
    }
}