[Swift]LeetCode551. 学生出勤纪录 I | Student Attendance Record I

★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★
➤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/9841699.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！
★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★

热烈欢迎，请直接点击！！！

进入博主App Store主页，下载使用各个作品！！！

注：博主将坚持每月上线一个新app！！！

You are given a string representing an attendance record for a student. The record only contains the following three characters:

1. 'A' : Absent.
2. 'L' : Late.
3. 'P' : Present.

 A student could be rewarded if his attendance record doesn't contain more than one 'A' (absent) or more than two continuous 'L' (late).

You need to return whether the student could be rewarded according to his attendance record.

Example 1:

Input: "PPALLP"
Output: True

 Example 2:

Input: "PPALLL"
Output: False

---

给定一个字符串来代表一个学生的出勤纪录，这个纪录仅包含以下三个字符：

1. 'A' : Absent，缺勤
2. 'L' : Late，迟到
3. 'P' : Present，到场

如果一个学生的出勤纪录中不超过一个'A'(缺勤)并且不超过两个连续的'L'(迟到),那么这个学生会被奖赏。

你需要根据这个学生的出勤纪录判断他是否会被奖赏。

示例 1:

输入: "PPALLP"
输出: True

示例 2:

输入: "PPALLL"
输出: False

---

8ms

class Solution {
    func checkRecord(_ s: String) -> Bool {
         if checkOnlyA(s) && checkContinuousL(s)
        {
            return true
        }
        return false
    }
    
    //检查不超过一个'A'(缺勤)
    func checkOnlyA(_ s: String) -> Bool
    {
        var count:Int = 0
        for char in s.characters
        {
            if char == "A"
            {
                count += 1
            }
            if count > 1
            {
                return false
            }
        }
        return true
    }
    
    //检查不超过两个连续的'L'(迟到)
    func checkContinuousL(_ s: String) -> Bool
    {
        if s.count < 3 {return true}
        for i in 0...s.count-3
        {
            var char1:Character = s[s.index(s.startIndex,offsetBy: i)]
            var char2:Character = s[s.index(s.startIndex,offsetBy: i + 1)]
            var char3:Character = s[s.index(s.startIndex,offsetBy: i + 2)]
            if char1 == "L" && char1 == char2 && char2 == char3
            {
                return false
            }
        }
        return true
    }
}

---

8ma

class Solution {
    func checkRecord(_ s: String) -> Bool {
        var lateCounter = 0
        var absentCounter = 0
        for character in s {
            if character == "L" {
                lateCounter += 1
                if lateCounter > 2 {
                    return false
                }
                continue
            }
        
            lateCounter = 0
            
            if character == "A" {
                absentCounter += 1
                if absentCounter > 1 {
                    return false
                }
            }            
        }
        return true
    }
}

---

12ms

class Solution {
    func checkRecord(_ s: String) -> Bool {
        let sArray = Array(s)
        var absentCount = 0
        var latenessCount = 0
        for i in 0..<sArray.count {
            if sArray[i] == Character("L") {
                latenessCount += 1
                
                if latenessCount > 2 {
                    return false
                }
            } else {
                latenessCount = 0

                if sArray[i] == Character("A") {
                    absentCount += 1
                }
            }
        }
        
        return absentCount < 2
    }
}

---

16ms

class Solution {
    func checkRecord(_ s: String) -> Bool {
      var numberOfAbsenses = 0
      var contigousTardies = 0
      
      for char in s {
        switch String(char) {
          case "A":
            numberOfAbsenses = numberOfAbsenses + 1
            contigousTardies = 0
            if numberOfAbsenses == 2 {
              return false
            }
          case "L":
            contigousTardies = contigousTardies + 1
          if contigousTardies == 3 {
              return false
            }
          default: 
            contigousTardies = 0
            continue
        }
      }
      return true
    }
}

---

20ms

class Solution {
    func checkRecord(_ s: String) -> Bool {
        var aCount = 0
        var lCount = 0
        for char in s {
            if char == "A" {
                aCount += 1
                if aCount == 2 {
                    return false
                }
                
                lCount = 0
            } else if char == "L" {
                lCount += 1
                if lCount == 3 {
                    return false
                }
            } else if char == "P" {
                lCount = 0
            }
        }

        return true
    }
}

---

24ms

class Solution {
    func checkRecord(_ s: String) -> Bool {
        var absentCount = 0
        var temp: Character
        
        for i in 0..<s.count {
            temp = s[s.index(s.startIndex, offsetBy: i)]
            if temp == "A" {
                absentCount += 1
                if absentCount >= 2 { return false }
            }
            else if  (s.count > (i + 2) && temp == "L" 
                    && s[s.index(s.startIndex, offsetBy: i + 1)] == "L"
                    && s[s.index(s.startIndex, offsetBy: i + 2)] == "L") {
                    return false
                }
        }
        
        return true
    }
}