LeetCode 551. Student Attendance Record I (C++版）

题目描述：

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

1. 'A' : Absent. 
2. 'L' : Late.
3. '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

思路分析：

判断为false的情况有两种：一是A的个数超过1个，二是连续的L的个数超过2个。

1）判断A的个数是否超过1个，使用一个count变量计数即可。

2）怎么判断连续的L的个数是否超过2个呢？——同样，借助一个count变量计算连续的L的个数，遍历字符串，每当遇到不是L的字符时，就置该值为0；每当遇到L时，计数就加1。若超过2，则返回false。

代码：

class Solution {public:    bool checkRecord(string s) {        int len = s.length();        if(len == 1 || len == 0) return true;                int cntA = 0;        int cnt_continous_L = 0;            for(int i = 0; i < len; i ++) {            if(s[i] == 'A') {                cntA ++;                if(cntA > 1) return false;                cnt_continous_L = 0;            }            else if(s[i] == 'L') {                cnt_continous_L ++;                if(cnt_continous_L > 2) return false;            }            else{                cnt_continous_L = 0;            }                    }        return true;    }};