LeetCode 第 125 題 (Valid Palindrome)

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.

這個問題也比較簡單，設定頭尾兩個指標。從兩邊向中間比較，發現不一樣的就退出。如果兩個指標指到同一個位置了還沒發現不一樣那就是正確的。

class Solution {
public:
    bool isPalindrome(string s) {
        string::const_iterator head = s.cbegin();
        string::const_iterator tail = s.cend();

        while(head < tail)
        {
            if(!isalnum(*head))
            {
                head ++;
                continue;
            }
            if(!isalnum(*tail))
            {
                tail --;
                continue;
            }
            if( tolower(*head) != tolower(*tail) )
            {
                return false;
            }
            head ++;
            tail --;
        }
        return true;
    }
};

如果不習慣用 iterator，還可以寫成 C 風格的。

class Solution {
public:
    bool isPalindrome(string s)
    {
        int n1 = 0, n2 = s.length() - 1;
        while(n1 < n2)
        {
            if(!isalnum(s[n1]))
            {
                n1 ++;
                continue;
            }
            if(!isalnum(s[n2]))
            {
                n2 --;
                continue;
            }
            if( tolower(s[n2]) != tolower(s[n1]) )
            {
                return false;
            }
            n1 ++;
            n2 --;
        }
        return true;
    }
};