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

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;
    }
};