125. Valid Palindrome(easy)

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Easy

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Given a string, determine if it is a palindrome, considering only alphanumeric characters and ignoring cases.

Note: For the purpose of this problem, we define empty string as valid palindrome.

Example 1:

Input: "A man, a plan, a canal: Panama"
Output: true

Example 2:

Input: "race a car"
Output: false

C++:

/*
 * @Author: SourDumplings
 * @Link: https://github.com/SourDumplings/
 * @Email: [email protected]
 * @Description: https://leetcode.com/problems/valid-palindrome/
 * @Date: 2019-03-20 18:31:23
 */

class Solution
{
  public:
    bool isPalindrome(string s)
    {
        int l = s.length();
        string str;
        str.resize(l);
        int j = 0;
        for (int i = 0; i < l; i++)
        {
            if (isalnum(s[i]))
            {
                str[j++] = tolower(s[i]);
            }
        }
        for (int i = 0; i < j / 2; i++)
        {
            if (str[i] != str[j - i - 1])
            {
                return false;
            }
        }
        return true;
    }
};

Java:

/*
 * @Author: SourDumplings
 * @Link: https://github.com/SourDumplings/
 * @Email: [email protected]
 * @Description: https://leetcode.com/problems/valid-palindrome/
 * @Date: 2019-03-20 18:43:18
 */

class Solution
{
    public boolean isPalindrome(String s)
    {
        StringBuffer str = new StringBuffer();
        int l = s.length();
        for (int i = 0; i < l; i++)
        {
            char c = s.charAt(i);
            if (Character.isAlphabetic(c) || Character.isDigit(c))
            {
                str.append(Character.toLowerCase(c));
            }
        }
        int ll = str.length();
        for (int i = 0; i < ll / 2; ++i)
        {
            if (str.charAt(i) != str.charAt(ll - i - 1))
            {
                return false;
            }
        }
        return true;
    }
}

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转载自blog.csdn.net/SourDumplings/article/details/88696847