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algorithm notes: leetcode#551 student attendance record i

admin 11月 12, 2020 0

Problem

Solution

Initial thoughts

Python implementation

class :
    def checkRecord(self, s):
        """
        :type s: str
        :rtype: bool
        """
        has_absent=False
        for i, c in enumerate(s):
            if c == 'A':
                if has_absent:
                    return False
                has_absent=True
            if i>1 and c=='L' and s[i-1]=='L' and s[i-2]=='L':
                return False
        return True

Java implementation

class  {
    public boolean checkRecord(String s) {
        boolean hasAbsent = false;
        for(int i = 0; i < s.length(); i++){
            if(s.charAt(i)=='A'){
                if(hasAbsent){ return false; }
                hasAbsent = true;
            }
            if(i>1 && s.charAt(i)=='L' && s.charAt(i-1)=='L' && s.charAt(i-2)=='L'){
                return false;
            }
        }
        return true;
    }
}

Time complexity

O(n).

Space complexity

O(1).

Links

551. Student Attendance Record I
(中文版) 算法笔记: 力扣#551 学生出勤记录 I