[LintCode] Valid Palindrome 验证回文字符串

判断回文字符串

最新推荐文章于 2024-01-04 12:53:14 发布

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最新推荐文章于 2024-01-04 12:53:14 发布

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原文链接：http://www.cnblogs.com/grandyang/p/5731347.html

本文介绍了一种算法，用于判断一个字符串是否为回文，只考虑字母数字字符并忽略大小写。提供了两种C++实现方法，均采用双指针策略，分别自定义了辅助函数和使用STL函数。

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

Notice

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.

Have you met this question in a real interview?

Example

"A man, a plan, a canal: Panama" is a palindrome.

"race a car" is not a palindrome.

Challenge

O(n) time without extra memory.

LeetCode上的原题，请参见我之前的博客Valid Palindrome。

解法一：

class Solution {
public:
    /**
     * @param s A string
     * @return Whether the string is a valid palindrome
     */
    bool isPalindrome(string& s) {
        int left = 0, right = s.size() - 1;
        while (left < right) {
            if (!isAlNum(s[left])) ++left;
            else if (!isAlNum(s[right])) --right;
            else if ((s[left] + 32 - 'a') % 32 != (s[right] + 32 - 'a') % 32) return false;
            else {
                ++left; --right;
            }
        }
        return true;
    }
    bool isAlNum(char c) {
        if (c >= 'a' && c <= 'z') return true;
        if (c >= 'A' && c <= 'Z') return true;
        if (c >= '0' && c <= '9') return true;
        return false;
    }
};

解法二：

class Solution {
public:
    /**
     * @param s A string
     * @return Whether the string is a valid palindrome
     */
    bool isPalindrome(string& s) {
        int left = 0, right = s.size() - 1;
        while (left < right) {
            if (!isalnum(s[left])) ++left;
            else if (!isalnum(s[right])) --right;
            else if ((s[left] + 32 - 'a') % 32 != (s[right] + 32 - 'a') % 32) return false;
            else {
                ++left; --right;
            }
        }
        return true;
    }
};