[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:

'A' : Absent.
'L' : Late.
'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：

'A' : Absent，缺勤
'L' : Late，迟到
'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 }

转载于:https://www.cnblogs.com/strengthen/p/9841699.html

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