[LeetCode]--125. 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.

可以忽略除数字和字母以外的标点符号。

 public boolean isPalindrome(String s) {
        if (s.length() == 0)
            return true;
        int i = 0, j = s.length() - 1;
        while (i < j) {
            while (i < j
                    && !((s.charAt(i) >= 'a' && s.charAt(i) <= 'z')
                            || (s.charAt(i) >= 'A' && s.charAt(i) <= 'Z') || (s
                            .charAt(i) >= '0' && s.charAt(i) <= '9')))
                i++;
            while (i < j
                    && !((s.charAt(j) >= 'a' && s.charAt(j) <= 'z')
                            || (s.charAt(j) >= 'A' && s.charAt(j) <= 'Z') || (s
                            .charAt(j) >= '0' && s.charAt(j) <= '9'))) {
                System.out.println("--" + j);
                j--;
            }
            if (!(s.charAt(i) + "").equalsIgnoreCase(s.charAt(j) + "")) {
                return false;
            }
            i++;
            j--;
        }
        return true;
    }

这样写通过了。下面学习下别人的。

public class Solution {
    public boolean isPalindrome(String s) {
        if (s == null || s.length() == 0) {
            return true;
        }

        int front = 0;
        int end = s.length() - 1;
        while (front < end) {
            while (front < s.length() && !isvalid(s.charAt(front))){ // nead to check range of a/b
                front++;
            }

            if (front == s.length()) { // for emtpy string “.,,,”     
                return true; 
            }           

            while (end >= 0 && ! isvalid(s.charAt(end))) { // same here, need to check border of a,b
                end--;
            }

            if (Character.toLowerCase(s.charAt(front)) != Character.toLowerCase(s.charAt(end))) {
                break;
            } else {
                front++;
                end--;
            }
        }

        return end <= front; 
    }

    private boolean isvalid (char c) {
        return Character.isLetter(c) || Character.isDigit(c);
    }
}