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