题目地址：https://leetcode.com/problems/student-attendance-record-i/#/descriptionopen in new window
题目描述
Youare 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).
Youneed 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
题目大意
如果A的次数不超过1次，连续L的次数不超过2次，那一定是优秀的同学，绝对是优秀的同学。
判断一个学生是不是优秀的同学。
解题方法
正则表达式
一直错的原因是没看清题，题目说的是不超过一个A或者两个连续的L，直接求解有点麻烦，可以用正则表达式。 .*匹配的是任意字符重复0次或者任意次，一行代码搞定。
public class Solution {
public boolean checkRecord(String s) {
return !s.matches(".*A.*A.*") && !s.matches(".*LLL.*");
}
}
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python写法如下，注意的是需要把正则表达式写在第一个参数，字符串s作为第二个参数。
class Solution:
def checkRecord(self, s):
"""
:type s: str
:rtype: bool
"""
return not re.match(".*A.*A.*", s) and not re.match(".*LLL.*", s)
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统计
直接数A的个数不能超过1，连续的L的个数不能多于2即可。代码还很简单高效的，超过了98%的提交。
class Solution:
def checkRecord(self, s):
"""
:type s: str
:rtype: bool
"""
N = len(s)
s = s + "P"
countA = 0
countL = 0
for i in range(N):
if s[i] == "A":
countA += 1
if countA > 1:
return False
if s[i] == "L":
countL += 1
if countL >= 3:
return False
else:
countL = 0
return True
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