[Swift]LeetCode334. 递增的三元子序列 | Increasing Triplet Subsequence

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Given an unsorted array return whether an increasing subsequence of length 3 exists or not in the array.

Formally the function should:

Return true if there exists i, j, k 
such that arr[i] < arr[j] < arr[k] given 0 ≤ i < j < k ≤ n-1 else return false.

Note: Your algorithm should run in O(n) time complexity and O(1) space complexity.

Example 1:

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

Example 2:

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

---

给定一个未排序的数组，判断这个数组中是否存在长度为 3 的递增子序列。

数学表达式如下:

如果存在这样的 i, j, k,  且满足 0 ≤ i < j < k ≤ n-1，
使得 arr[i] < arr[j] < arr[k] ，返回 true ; 否则返回 false 。

说明: 要求算法的时间复杂度为 O(n)，空间复杂度为 O(1) 。

示例 1:

输入: [1,2,3,4,5]
输出: true

示例 2:

输入: [5,4,3,2,1]
输出: false

---

16ms

class Solution {
    func increasingTriplet(_ nums: [Int]) -> Bool {
        guard nums.count >= 3 else {
            return false
        }
        
        var first = Int.max
        var second = Int.max
        
        for num in nums {
            if num <= first {
                first = num
            } else if num <= second {
                second = num
            } else {
                return true
            }
        }
        return false
    }
}

---

20ms

class Solution {
       func increasingTriplet(_ nums: [Int]) -> Bool {
   
    guard nums.count >= 3 else {
      
      return false
    }
    
    var minvalue = nums[0]
    
    var middle = Int.max
    
    var minvalue1 = nums[0]
    
    for i in 1..<(nums.count - 1) {
      
      if nums[i] <= minvalue1 {
        
        if middle != Int.max {
          
          minvalue1 = nums[i]
        }
        else {
          
          minvalue = nums[i]
          
          minvalue1 = minvalue
          
          middle = Int.max
        }
      }
      else if nums[i] <= middle {
        
        middle = nums[i]
        
        minvalue = minvalue1
      }
      else {
        
        return true
      }
    }
    
    return minvalue < middle && middle < nums.last!
  }
}

---

48ms

class Solution {
    func increasingTriplet(_ nums: [Int]) -> Bool {
        if nums.count < 3 {
            return false
        }
        var minI = nums[0]
        var bigMin : Int? = nil
        
        for i in 1..<nums.count {
            if bigMin != nil && nums[i] > bigMin! {
                return true
            }
            
            if nums[i] < minI {
                minI = nums[i]
            }
            if nums[i] > minI {
                if bigMin == nil {
                    bigMin = nums[i]
                }else {
                    bigMin = min(nums[i], bigMin!)
                }
            }
        }
        
        return false
    }
}

---

56ms

class Solution {
    func increasingTriplet(_ nums: [Int]) -> Bool {
        var minInt:Int = Int.max
        var maxInt:Int = Int.max
        for item in nums {
            if minInt >= item {
                minInt = item
            }else if maxInt >= item {
                maxInt = item
            }else{
                return true
            }
        }
        return false
    }
}

---

56ms

class Solution {
    func increasingTriplet(_ nums: [Int]) -> Bool {
        guard nums.count > 2 else {
            return false
        }

        var n1 = Int.max
        var n2 = Int.max - 1

        for n in nums {
            if n <= n1 {
                n1 = n
            } else if n < n2 {
                n2 = n
            } else if n1 < n2 && n2 < n {
                return true
            }
        }
        
        return false
    }
}