[Swift]LeetCode414. 第三大的数 | Third Maximum Number

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Given a non-empty array of integers, return the third maximum number in this array. If it does not exist, return the maximum number. The time complexity must be in O(n).

Example 1:

Input: [3, 2, 1]

Output: 1

Explanation: The third maximum is 1.

Example 2:

Input: [1, 2]

Output: 2

Explanation: The third maximum does not exist, so the maximum (2) is returned instead.

Example 3:

Input: [2, 2, 3, 1]

Output: 1

Explanation: Note that the third maximum here means the third maximum distinct number.
Both numbers with value 2 are both considered as second maximum.

---

给定一个非空数组，返回此数组中第三大的数。如果不存在，则返回数组中最大的数。要求算法时间复杂度必须是O(n)。

示例 1:

输入: [3, 2, 1]

输出: 1

解释: 第三大的数是 1.

示例 2:

输入: [1, 2]

输出: 2

解释: 第三大的数不存在, 所以返回最大的数 2 .

示例 3:

输入: [2, 2, 3, 1]

输出: 1

解释: 注意，要求返回第三大的数，是指第三大且唯一出现的数。
存在两个值为2的数，它们都排第二。

---

12ms:用元组记录数值

class Solution {
    func thirdMax(_ nums: [Int]) -> Int {
        var (num0,num1,num2) = (Int.min,Int.min,Int.min)
        for num in nums {
            if num == num0 || num == num1 || num == num2 {
                continue
            }
            if num > num0 {
                (num0,num1,num2) = (num,num0,num1)
                continue
            }
            if num > num1 {
                (num1,num2) = (num,num1)
                continue
            }
            if num > num2 {
                num2 = num
            }
        }
        return num2 == Int.min ? num0 : num2
    }
}

---

16ms

class Solution {
    func thirdMax(_ nums: [Int]) -> Int {
        var firstMax = nums[0]
        var secondMax: Int?
        var thirdMax: Int?
        
        // get max
        for num in nums {
            firstMax = max(firstMax, num)
        }
        // get 2nd max
        for num in nums {
            if num < firstMax {
                if let curSecondMax = secondMax {
                    secondMax = max(curSecondMax, num)
                } else {
                    secondMax = num
                }
            }
        }
        
        if secondMax == nil {
            return firstMax
        }
        
        for num in nums {
            if num < secondMax! {
                if let curThirdMax = thirdMax {
                    thirdMax = max(curThirdMax, num)
                } else {
                    thirdMax = num
                }
            }
        }
        
        return thirdMax ?? firstMax
    }
}

---

16ms

class Solution {
    func thirdMax(_ nums: [Int]) -> Int {
        var firstMax = Int.min
        var secondMax = Int.min
        var thirdMax = Int.min
        var count = 0

        for num in nums {
            if num > firstMax {
                (thirdMax, secondMax, firstMax) = (secondMax, firstMax, num)
                count += 1
            } else if num != firstMax && num > secondMax {
                (thirdMax, secondMax) = (secondMax, num)
                count += 1
            } else if num != firstMax && num != secondMax && num > thirdMax {
                thirdMax = num
                count += 1
            }
        }

        if count < 3 {
            return firstMax
        }

        return thirdMax
    }
}

---

20ms

class Solution {
    func thirdMax(_ nums: [Int]) -> Int {
        var max = nums[0]
        var second: Int?
        var third: Int?
        
        for i in 1..<nums.count {
            if nums[i] < max {
                if second == nil {
                    second = nums[i]
                } else {
                    if second! < nums[i] {
                        third = second!
                        second = nums[i]
                    } else if second! > nums[i] {
                        if third == nil {
                            third = nums[i]
                        } else {
                            if nums[i] > third! {
                                third = nums[i]
                            }
                        }
                        
                    }
                }
                
            } else if nums[i] > max {
                third = second
                second = max
                max = nums[i]
            }
        }
        
        if third != nil {
            return third!
        }
        
        return max
    }
}

---

24ms

class Solution {
    func thirdMax(_ nums: [Int]) -> Int {
        var setNum = Set<Int>(nums)
        var queue = Queue<Int>(size: 3)
        
        for num in setNum {
            if queue.count < 3 {
                queue.push(value: num)
            } else if let min = queue.min() {
                if num > min {
                    queue.pop()
                    queue.push(value: num)
                }
            } else {
                queue.push(value: num)
            }
        }
        
        if queue.count < 3 {
            while queue.count != 1 {
                queue.pop()
            }
        }

        return queue.min() ?? 0       
    }    
}

class Queue<T: Comparable> {
    private var data = [T]()
    private let size: Int
    var count: Int {
        return data.count
    }
    init(size: Int) {
        self.size = size
    }
    
    func push(value: T) {
        if data.isEmpty {
            data.append(value)
        } else if data.count + 1 > size {
            if let last = data.first {
                if value < last {
                    data.removeFirst()
                    push(value: value)
                }
            }
        } else {
            var toInsertAt = -1
            for (index, dataValue) in data.enumerated() {
                if value > dataValue {
                    toInsertAt = index
                    break
                }
            }
            if toInsertAt == -1 {
                data.append(value)
            } else {
                data.insert(value, at: toInsertAt)
            }
        }
    }
    
    @discardableResult
    func pop() -> T? {
        return data.popLast()
    }
    
    @discardableResult
    func min() -> T? {
        if data.isEmpty {
            return nil
        } else {
            return data.last
        }
    }   
}