【Leetcode】Valid Palindrome

最新推荐文章于 2019-07-29 08:28:36 发布

yyz20002008

最新推荐文章于 2019-07-29 08:28:36 发布

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文章标签： leetcode Java String char

本文提供了一种简单的方法来判断一个字符串是否为回文，仅考虑字母数字字符并忽略大小写。通过双指针技术从两端向中间遍历，跳过非字母数字字符进行比较。

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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.

Java:

http://blog.youkuaiyun.com/linhuanmars/article/details/22775045

还好比较简单

public class Solution {
    public boolean isPalindrome(String s) {
        if(s==null||s.length()==0) return true;
        int l=0;
        int r=s.length()-1;
        while(l<r)
        {
            if(!isValid(s.charAt(l)))
            {
                l++;
                continue;
            }
            if(!isValid(s.charAt(r)))
            {
                r--;
                continue;
            }
            if(!isSame(s.charAt(l),s.charAt(r)))
            {
                return false;
            }
            l++;
            r--;
            
        }
        return true;
    }
    private boolean isValid(char c)
    {
        if(c>='a'&&c<='z'||c>='A'&&c<='Z'||c>='0'&&c<='9') return true;
        return false; 
    }
    private boolean isSame(char c1, char c2)
    {
        if(c1>='A'&&c1<='Z')
        c1=(char)(c1-'A'+'a');
        if(c2>='A'&&c2<='Z')
        c2=(char)(c2-'A'+'a');
        return c1==c2;
    }
}