Valid Palindrome

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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分类专栏： Leedcode

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本文介绍了如何通过算法判断一个字符串是否为回文，只考虑字母和数字字符，并且忽略了大小写的影响。包括了字符串转换、回文判断的详细步骤。

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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

思路：

1.只剩下字母

2.转化为小写

3.回文比较

#include <iostream>
using namespace std;

class Solution{
public:
	//转换大小写
	string low(string s)
	{
		for(int i=0;i<s.length();i++)
		{
			if(s[i]>='A' && s[i]<='Z') s[i] = s[i] - 'A' + 'a';
		}
		return s;
	}

	//判断是否是字母或者数字
	bool isAlpha(char c)
	{
		if((c>='a' && c<='z') || (c>='0' && c<='9')) return true;
		return false;
	}

	bool isPalindrome(string s)
	{
		//转换大小写
		s=low(s);

		//左右夹逼,判断是不是回文字符串
		int left = 0;
		int right = s.length() -1;
		while(left < right)
		{
			while(left<s.length()&&!isAlpha(s[left]))
			{
				left ++;
			}
			while(right >=0 && !isAlpha(s[right]))
			{
				right --;
			}
			if(left>right || left>=s.length()||right <0) break;//异常处理，比如空字符串
			if(s[left]!=s[right])
			{
				return false;
			}
			left++;
			right--;
		}
		return true;
	}
};

void main()
{
	string s="";
	Solution *sol = new Solution();
	bool flag = sol->isPalindrome(s);
	printf("The string is %s Palindrome.\n",flag?"true":"false");
	delete sol;
	
}