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

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    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
     1 class Solution {
     2     func checkRecord(_ s: String) -> Bool {
     3          if checkOnlyA(s) && checkContinuousL(s)
     4         {
     5             return true
     6         }
     7         return false
     8     }
     9     
    10     //检查不超过一个'A'(缺勤)
    11     func checkOnlyA(_ s: String) -> Bool
    12     {
    13         var count:Int = 0
    14         for char in s.characters
    15         {
    16             if char == "A"
    17             {
    18                 count += 1
    19             }
    20             if count > 1
    21             {
    22                 return false
    23             }
    24         }
    25         return true
    26     }
    27     
    28     //检查不超过两个连续的'L'(迟到)
    29     func checkContinuousL(_ s: String) -> Bool
    30     {
    31         if s.count < 3 {return true}
    32         for i in 0...s.count-3
    33         {
    34             var char1:Character = s[s.index(s.startIndex,offsetBy: i)]
    35             var char2:Character = s[s.index(s.startIndex,offsetBy: i + 1)]
    36             var char3:Character = s[s.index(s.startIndex,offsetBy: i + 2)]
    37             if char1 == "L" && char1 == char2 && char2 == char3
    38             {
    39                 return false
    40             }
    41         }
    42         return true
    43     }
    44 }

    8ma

     1 class Solution {
     2     func checkRecord(_ s: String) -> Bool {
     3         var lateCounter = 0
     4         var absentCounter = 0
     5         for character in s {
     6             if character == "L" {
     7                 lateCounter += 1
     8                 if lateCounter > 2 {
     9                     return false
    10                 }
    11                 continue
    12             }
    13         
    14             lateCounter = 0
    15             
    16             if character == "A" {
    17                 absentCounter += 1
    18                 if absentCounter > 1 {
    19                     return false
    20                 }
    21             }            
    22         }
    23         return true
    24     }
    25 }

    12ms

     1 class Solution {
     2     func checkRecord(_ s: String) -> Bool {
     3         let sArray = Array(s)
     4         var absentCount = 0
     5         var latenessCount = 0
     6         for i in 0..<sArray.count {
     7             if sArray[i] == Character("L") {
     8                 latenessCount += 1
     9                 
    10                 if latenessCount > 2 {
    11                     return false
    12                 }
    13             } else {
    14                 latenessCount = 0
    15 
    16                 if sArray[i] == Character("A") {
    17                     absentCount += 1
    18                 }
    19             }
    20         }
    21         
    22         return absentCount < 2
    23     }
    24 }

    16ms

     1 class Solution {
     2     func checkRecord(_ s: String) -> Bool {
     3       var numberOfAbsenses = 0
     4       var contigousTardies = 0
     5       
     6       for char in s {
     7         switch String(char) {
     8           case "A":
     9             numberOfAbsenses = numberOfAbsenses + 1
    10             contigousTardies = 0
    11             if numberOfAbsenses == 2 {
    12               return false
    13             }
    14           case "L":
    15             contigousTardies = contigousTardies + 1
    16           if contigousTardies == 3 {
    17               return false
    18             }
    19           default: 
    20             contigousTardies = 0
    21             continue
    22         }
    23       }
    24       return true
    25     }
    26 }

    20ms

     1 class Solution {
     2     func checkRecord(_ s: String) -> Bool {
     3         var aCount = 0
     4         var lCount = 0
     5         for char in s {
     6             if char == "A" {
     7                 aCount += 1
     8                 if aCount == 2 {
     9                     return false
    10                 }
    11                 
    12                 lCount = 0
    13             } else if char == "L" {
    14                 lCount += 1
    15                 if lCount == 3 {
    16                     return false
    17                 }
    18             } else if char == "P" {
    19                 lCount = 0
    20             }
    21         }
    22 
    23         return true
    24     }
    25 }

    24ms

     1 class Solution {
     2     func checkRecord(_ s: String) -> Bool {
     3         var absentCount = 0
     4         var temp: Character
     5         
     6         for i in 0..<s.count {
     7             temp = s[s.index(s.startIndex, offsetBy: i)]
     8             if temp == "A" {
     9                 absentCount += 1
    10                 if absentCount >= 2 { return false }
    11             }
    12             else if  (s.count > (i + 2) && temp == "L" 
    13                     && s[s.index(s.startIndex, offsetBy: i + 1)] == "L"
    14                     && s[s.index(s.startIndex, offsetBy: i + 2)] == "L") {
    15                     return false
    16                 }
    17         }
    18         
    19         return true
    20     }
    21 }
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  • 原文地址:https://www.cnblogs.com/strengthen/p/9841699.html
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