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551. Student Attendance Record I

时间:2019-03-08 12:44:22      阅读:177      评论:0      收藏:0      [点我收藏+]

标签:c++   following   inpu   repr   wing   out   bsp   pre   OWIN   

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

 

Approach #1: Brut force. [C++]

class Solution {
public:
    bool checkRecord(string s) {
        int numOfA = 0, numOfL = 0;
        for (int i = 0; i < s.length(); ++i) {
            if (s[i] == ‘A‘) numOfA++;
            if (s[i] == ‘L‘) numOfL++;
            else numOfL = 0;
            if (numOfA > 1 || numOfL > 2) return false; 
        }

        return true;
    }
};

  

 

551. Student Attendance Record I

标签:c++   following   inpu   repr   wing   out   bsp   pre   OWIN   

原文地址:https://www.cnblogs.com/ruruozhenhao/p/10495000.html

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