valid-palindrome

【题目描述】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.

【解题思路】验证回文字符串是比较常见的问题，所谓回文，就是一个正读和反读都一样的字符串，比如“level”或者“noon”等等就是回文串。但是这里，加入了空格和非字母数字的字符，增加了些难度，但其实原理还是很简单：只需要建立两个指针，left和right, 分别从字符的开头和结尾处开始遍历整个字符串，如果遇到非字母数字的字符就跳过，继续往下找，直到找到下一个字母数字或者结束遍历，如果遇到大写字母，就将其转为小写。等左右指针都找到字母数字时，比较这两个字符，若相等，则继续比较下面两个分别找到的字母数字，若不相等，直接返回false.

【考查内容】递归

class Solution {
public:
    bool isPalindrome(string s) {
       int left = 0, right = s.size() - 1 ;
        while (left < right) {
            if (!isAlphaNum(s[left])) ++left;
            else if (!isAlphaNum(s[right])) --right;
            else if ((s[left] + 32 - 'a') %32 != (s[right] + 32 - 'a') % 32) return false;
            else {
                ++left; --right;
            }
        }
        return true;
    }
    bool isAlphaNum(char &ch) {
        if (ch >= 'a' && ch <= 'z') return true;
        if (ch >= 'A' && ch <= 'Z') return true;
        if (ch >= '0' && ch <= '9') return true;
        return false;
    }
};