551. Student Attendance Record I

QUESTION:

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

EXPLANATION:

其实我一开始想到的就是遍历数组，然后筛选出acount和lcount数量，最后再进行检查。然后在自己写测试用例的时候发现，如果后面很多都是p的话，其实会造成很多的性能浪费，所以，在for循环之后进行了条件判断。这样在如果string不是很长的时候或许是有性能浪费的。

如果是正常情况的话，学生其实应该是p的多，大部分应该都是true可以获得嘉奖的，但是因为是leetcode，所以很明显应该是false的多，所以最后的耗时是9ms。

然后看了一下discuss中的，发现其他人都是使用的正则表达式，正则表达式可以让代码很容易被看懂，但是我也测试了下性能，发现需要24ms，所以孰轻孰重还是看你取舍了。

SOLUTION:

public class Solution {
    public boolean checkRecord(String s) {
        char[] chars = s.toCharArray();
        int aCount =0,lCount = 0;
        for(char ch : chars){
            switch (ch) {
                case 'A':
                    aCount ++;
                    lCount = 0;
                    break;
                case 'P':
                    lCount = 0;
                    break;
                case 'L':
                    lCount++;
                    break;
            }
            if(aCount>1 || lCount>2)
                return false;
        }
        return true;
    }
}


public class Solution {
    public boolean checkRecord(String s) {
        return !s.matches(".*LLL.*|.*A.*A.*");
    }
}
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