描述

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

分析

P不用管，A和L分别用两个变量aCount和lCount来记录。对于‘A’，每遇到一个就让aCount加一，大于等于2时直接退出主循环；对于‘L’，一旦遇到就进入一个循环，若下一个元素也为‘L’，则将lCount加1，直到不是‘L’或到了string的末尾为止。由于循环中多加了一次i，因此退出循环后要减一次i。如果此时lCount小于等于2，就置为0，否则退出主循环。

退出主循环后，判断lCount和aCount的个数，若符合条件则返回false，否则返回true。

代码如下：

class Solution {

public:

    bool checkRecord(string s) {

        int aCount = 0;

        int lCount = 0;

        for(int i = 0; i != s.size(); ++i){

            if(aCount > 1)break;

            if(s[i] == 'A')++aCount;

            if(s[i] == 'L'){

                while(s[i] == 'L' && i != s.size()){

                    ++lCount;

                    ++i;

                }

                --i;

                if(lCount <= 2)lCount = 0;

                else    break;

            }

        }

        if(aCount > 1 || lCount > 2)return false;

        return true;

    }

};

巴特西

leetcode解题报告（29）：Student Attendance Record I

描述

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