[LeetCode]40. Valid Palinadrome有效回文串

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.

解法1：先扫描一遍字符串，将除数字和字母外的字符扔掉，并且都转为小写；然后两个指针从两头向中间靠拢，一一比较对应字符是否相同即可。

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

解法2：第二个方法是两个指针从两头向中间靠拢，如果碰到非数字字母字符则跳过，然后一一比较即可。

class Solution {
public:
    bool isPalindrome(string s) {
        int n = s.size();
        int i = 0, j = n - 1;
        while (i <= j)
        {
            while (i < n && !isAlphanumeric(s[i])) ++i;
            while ( j >= 0 && !isAlphanumeric(s[j])) --j;
            if (i < n && j >= 0 && tolower(s[i]) != tolower(s[j])) return false;
            ++i; --j;
        }
        return true;
    }
private:
    bool isAlphanumeric(const char c)
    {
        if ((c >= '0' && c <= '9') || (c >= 'A' && c <= 'Z') || (c >= 'a' && c <= 'z'))
            return true;
        return false;
    }
};