Valid Palindrome

最新推荐文章于 2025-09-06 18:22:26 发布

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原文链接：http://www.cnblogs.com/kwill/archive/2013/02/27/2935945.html

文章标签：

#c/c++

本文介绍如何通过C/C++实现判断一个只考虑字母数字字符且忽略大小写的字符串是否为回文串的方法，包括特殊情况处理及代码实现。

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.

#include <stdlib.h>
#include <stdio.h>
#include <string>
using std::string;

class Solution {
    public:
        bool isAlphaNumber(char & c)
        {
            if (c >= '0' && c <= '9')
            {
                return true;
            }
            if (c >= 'a' && c <= 'z')
            {
                return true;
            }
            if (c >= 'A' && c <= 'Z')
            {
                c += 'a' - 'A';
                return true;
            }
            return false;
        }
        bool isPalindrome(string s) {
            // Start typing your C/C++ solution below
            //         // DO NOT write int main() function
            //                 
            if (s.empty())
            {
                return true;
            }
            int i = 0, j = s.size() - 1;
            char a,b;
            while(i < s.size() && j >= 0){
                a = s.at(i);
                b = s.at(j);
                if (!isAlphaNumber(a))
                {
                    i++;
                    continue;
                }
                if (!isAlphaNumber(b))
                {
                    j--;
                    continue;
                }
                if (a == b)
                {
                    i++;
                    j--;
                }else
                {
                    return false;
                }
            }
            //if (i + j != s.size())
            //{
            //    return false;
            //}else
            {
                return true;
            }
        }
};