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

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

Example 1:

Input: "A man, a plan, a canal: Panama"
Output: true

Example 2:

Input: "race a car"
Output: false

Constraints:

s consists only of printable ASCII characters.

solution

Time complexity : O(n)
Space complexity : O(1)

class Solution {
public:
    bool isPalindrome(string s) {
        for (int i=0, j=s.length()-1; i<j; ++i,--j) {
            while (i<j && !isalnum(s[i])) ++i;
            while (i<j && !isalnum(s[j])) --j;
            if (tolower(s[i]) != tolower(s[j]))
                return false;
        }
        return true;
    }
};

使用2-pointer，從頭、從尾，一路往中間檢查。
移動到是字母或是數字，統一轉換為小寫。
若途中有不一致，即為invalid palindrome。
直到 j < i (palindrome長度為偶)、或是 i == j (palindrome長度為奇，中間單獨一個字)，即結束。