C#LeetCode刷题之#551-学生出勤纪录 I​​​​​​​(Student Attendance Record I)

C#LeetCode刷题之#551-学生出勤纪录 I​​​​​​​(Student Attendance Record I)

问题

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。

输入: “PPALLP”

输出: True

输入: “PPALLL”

输出: False

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.

Input: “PPALLP”

Output: True

Input: “PPALLL”

Output: False

示例

public class Program {

    public static void Main(string[] args) {
        var s = "PPALLP";

        var res = CheckRecord(s);
        Console.WriteLine(res);

        s = "AAAA";

        res = CheckRecord2(s);
        Console.WriteLine(res);

        Console.ReadKey();
    }

    private static bool CheckRecord(string s) {
        //一些小技巧
        var boolA = (s.Length - s.Replace("A", "").Length) <= 1;
        var boolL = (s.Length == s.Replace("LLL", "").Length);
        return boolA && boolL;
    }

    private static bool CheckRecord2(string s) {
        //传统计数法
        var countA = 0;
        var countL = 0;
        foreach(var c in s) {
            if(c == 'A') ++countA;
            if(c == 'L') ++countL;
            else countL = 0;
            if(countA > 1 || countL > 2) return false;
        }
        return true;
    }

}

以上给出2种算法实现,以下是这个案例的输出结果:

True
False

分析:

因为使用了部分运行库,不能简单的认为 CheckRecord 的时间复杂度为 O(1)​ 。以上2种算法的时间复杂度均为 O(n)

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