[LeetCode]125. Valid Palindrome

题目描述：

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.

思路：

判断是否回文字符串。非字符、非数字不算，大小写不分。
首先去除非数字、非字符的标点等，存入另一个字符串（不需要存入另外的字符串，只要从头和尾开始比较就行了，遇到非字母跳过，相等则继续，直到i = j）
然后从头和尾开始比较，直到中间位置为止。如果都相等则返回true
i 和 j 表示字符串的开头和结尾位，如果不是合法的字符则i++,j--，如果合法则将它转化为小写字符，再比较，相等则比较下一位，否则退出。
最终返回true
注意while的用法，continue的用法
大小写均转化为小写是character.toLowerCase();

public class Solution125 {
    public boolean isPalindrome(String s) {
        if(s == null) return false;
        int i=0;
        int j = s.length()-1;
        while(i<j){
                if(!isAlphanumeric(s.charAt(i)))  {i++;continue;}
                if(!isAlphanumeric(s.charAt(j)))  {j--;continue;}
                if(Character.toLowerCase(s.charAt(i))!=Character.toLowerCase(s.charAt(j))) {
                    return false;
                }else {
                    i++;
                    j--;
                }
            }
        return true;
        }
        
    public boolean isAlphanumeric(char c){
        if((c >= 'A' && c <= 'Z') || (c >= 'a' && c <= 'z') || (c >= '0' && c <= '9'))
            {return true;}
        return false;
    }
    public static void main(String[] args) {
        // TODO Auto-generated method stub
        Solution125 solution125 = new Solution125();
        String s = "A man, a plan, a canal: Panama";
        if(solution125.isPalindrome(s)==true)
            System.out.println("1");
        else {
            System.out.println("0");
        }
    }

}