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： [cpp] 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： [cpp] 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中采用判斷刪除的方法，效率要高，且不占用另外字符串存儲（str）
