551. Student Attendance Record I【easy】
You are given a string representing an attendance record for a student. The record only contains the following three characters:
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
解法一：
1 class Solution { 2 public: 3 bool checkRecord(string s) { 4 int num_a = 0; 5 6 for (int i = 0; i < s.length(); ++i) { 7 if (s[i] == ‘A‘) { 8 if (++num_a > 1) { 9 return false; 10 } 11 } 12 13 if (i > 0 && i < s.length() - 1 14 && s[i] == ‘L‘ && s[i - 1] == ‘L‘ && s[i + 1] == ‘L‘) { 15 return false; 16 } 17 } 18 19 return true; 20 } 21 };
解法二：
1 public boolean checkRecord(String s) { 2 int countA=0; 3 int continuosL = 0; 4 int charA = ‘A‘; 5 int charL =‘L‘; 6 for(int i=0;i<s.length();i++){ 7 if(s.charAt(i) == charA){ 8 countA++; 9 continuosL = 0; 10 } 11 else if(s.charAt(i) == charL){ 12 continuosL++; 13 } 14 else{ 15 continuosL = 0; 16 } 17 if(countA >1 || continuosL > 2 ){ 18 return false; 19 } 20 } 21 return true; 22 23 }
参考@rakeshuky 的代码。
解法三：
1 public boolean checkRecord(String s) { 2 int acount=0; 3 int lcount=0; 4 for(char c : s.toCharArray()) { 5 if(c==‘L‘) { 6 lcount++; 7 } 8 else { 9 lcount=0; 10 if(c==‘A‘) { 11 acount++; 12 } 13 } 14 if(acount>1 || lcount>=3) { 15 return false; 16 } 17 } 18 return true; 19 }
参考@labs 的代码。
551. Student Attendance Record I【easy】
原文：http://www.cnblogs.com/abc-begin/p/7583324.html