【leetcode】Valid Palindrome

Question：

Given a string, determine if it is a palindrome, considering only alphanumeric characters and ignoring cases.

For example,
"A man, a plan, a canal: Panama" is a palindrome.
"race a car" is not a palindrome.

Note:
Have you consider that the string might be empty? This is a good question to ask during an interview.

For the purpose of this problem, we define empty string as valid palindrome.

Anwser 1：   

class Solution {public:    bool isStr(char &ch){        if(ch >= '0' && ch <= '9'){            return true;        } else if(ch >= 'a' && ch <= 'z'){            return true;        } else if(ch >= 'A' && ch <= 'Z'){           ch += 32;           return true;        }                 return false;    }        bool isPalindrome(string s) {        // Start typing your C/C++ solution below        // DO NOT write int main() function        int len = s.length();        if(len == 0){            return true;        }                string str = "";        for(int i = 0; i < len; i++){   // remove illegal char, such as "?" "/" ...            if(isStr(s[i])){                str += s[i];            }        }                len = str.length();        int mid = (len + 1) / 2;        for(int i = 0; i < mid; i++){            if(str[i] != str[len - 1 - i]){     // check front and end char                return false;            }        }        return true;    }};

Anwser 2：   

class Solution {public:    bool isStr(char &ch){        if(ch >= '0' && ch <= '9'){            return true;        } else if(ch >= 'a' && ch <= 'z'){            return true;        } else if(ch >= 'A' && ch <= 'Z'){           ch += 32;           return true;        }                 return false;    }        bool isPalindrome(string s) {        // Start typing your C/C++ solution below        // DO NOT write int main() function        int len = s.length();        if(len == 0){            return true;        }                int i = 0;        int j = len - 1;        while(i < j){            if(!isStr(s[i])) {                i++;            } else if(!isStr(s[j])) {                j--;            } else if(s[i++] != s[j--]) {                return false;            }        }                return true;    }};

注意點：

1） 字母數字判斷，注意字母中的字母大小寫轉換（把大寫加32全部轉化成小寫）

2） 方法2中採用判斷刪除的方法，效率要高，且不佔用另外字串儲存（str）