Leetcode: 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.

第二次做法：用API

public class Solution {
    public boolean isPalindrome(String s) {
        if (s == null) return false;
        if (s.length() == 0) return true;
        int l=0, r=s.length()-1;
        while (l < r) {
            while (l<r && !Character.isLetterOrDigit(s.charAt(l))) l++;
            while (l<r && !Character.isLetterOrDigit(s.charAt(r))) r--;
            if (l == r) break;
            if (Character.toLowerCase(s.charAt(l)) != Character.toLowerCase(s.charAt(r))) return false;
            else {
                l++;
                r--;
            }
        }
        return true;
    }
}

第一次做法：难度：79. 这道题是判断一个字符串是不是回文串。因为只是看一个字符串，算法还是比较简单，就是从两头出发，往中间走，进行两两匹配。这里面的小问题就是在这个题目要求中，只判断字母和数字类型的字符，其他字符直接跳过即可。因此我们要写一个函数判断他是不是合法字符，而且因为忽略大小写，我们在判断两个字符是不是相同的时候如果是大写，要转成相应的小写字母。这个算法从两边扫描，到中间相遇，只需要一次线性扫描，复杂度是O(n)，空间上是O(1)。

public class Solution {
    public boolean isPalindrome(String s) {
        if (s == null) return false;
        if (s.length() == 0) return true;
        int l=0, r=s.length()-1;
        while (l < r) {
            while (l<r && !isAlpha(s, l)) l++;
            while (l<r && !isAlpha(s, r)) r--;
            if (l == r) break;
            if (!isSame(s, l, r)) return false;
            else {
                l++;
                r--;
            }
        }
        return true;
    }
    
    public boolean isAlpha(String s, int i) {
        char c = s.charAt(i);
        if (c>='A'&&c<='Z' || c>='a'&&c<='z' || c>='0'&&c<='9') return true;
        return false;
    }
    
    public boolean isSame(String s, int l, int r) {
        char left = s.charAt(l);
        char right = s.charAt(r);
        if (left>='A' && left<='Z') left = (char)(left-'A'+'a');
        if (right>='A' && right<='Z') right = (char)(right-'A'+'a');
        return left==right;
    }
}

 28行和29行如果不强制转换类型，会出现possible loss of precision的错误