Valid Palindrome

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

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.

Challenge
O(n) time without extra memory.

字符串的回文判断问题，由于字符串可随机访问，故逐个比较首尾字符是否相等最为便利，即常见的『两根指针』技法。此题忽略大小写，并只考虑字母和数字字符。

C++：

class Solution {
public:
    /**
     * @param s A string
     * @return Whether the string is a valid palindrome
     */
    bool isPalindrome(string& s) {
        if (s.empty()) return true;

        int l = 0, r = s.size() - 1;
        while (l < r) {
            // find left alphanumeric character
            if (!isalnum(s[l])) {
                ++l;
                continue;
            }
            // find right alphanumeric character
            if (!isalnum(s[r])) {
                --r;
                continue;
            }
            // case insensitive compare
            if (tolower(s[l]) == tolower(s[r])) {
                ++l;
                --r;
            } else {
                return false;
            }
        }

        return true;
    }
};

JAVA：

public class Solution {
    /**
     * @param s A string
     * @return Whether the string is a valid palindrome
     */
    public boolean isPalindrome(String s) {
        if (s == null || s.isEmpty()) return true;

        int l = 0, r = s.length() - 1;
        while (l < r) {
            // find left alphanumeric character
            if (!Character.isLetterOrDigit(s.charAt(l))) {
                l++;
                continue;
            }
            // find right alphanumeric character
            if (!Character.isLetterOrDigit(s.charAt(r))) {
                r--;
                continue;
            }
            // case insensitive compare
            if (Character.toLowerCase(s.charAt(l)) == Character.toLowerCase(s.charAt(r))) {
                l++;
                r--;
            } else {
                return false;
            }
        }

        return true;
    }
}

源码分析

两步走：

1. 找到最左边和最右边的第一个合法字符(字母或者字符)
2. 一致转换为小写进行比较

字符的判断尽量使用语言提供的 API

复杂度分析

两根指针遍历一次，时间复杂度 O(n), 空间复杂度 O(1).