Leetcode Valid Palindrome

Valid Palindrome

Total Accepted:6669Total Submissions:31102My Submissions

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"isnota 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.

很简单的题目： 2星级。

考点：

1 会判断是否是字母和数字

2 知道什么是Palindrome

可以利用函数库判断是否为数字或者字母，程序如下：

bool isPalindrome(string s) 
	{
		for (int i = 0, j = s.size()-1; i < j; i++, j--)
		{
			while (i<j && (!isalpha(s[i]) && !isdigit(s[i]))) 
				i++;
			while (i<j && (!isalpha(s[j]) && !isdigit(s[j]))) 
				j--;
			
			if (i<j && tolower(s[i]) != tolower(s[j])) return false;
		}
		return true;
	}

自家写函数判断也很简单：

bool isPalindrome(string s) 
	{
		for (int i = 0, j = s.size()-1; i < j; i++, j--)
		{
			while (s[i] && (!isTestChar(s[i]))) i++;
			while (s[j] && (!isTestChar(s[j]))) j--;
			
			if (i<j && !isEqual(s[i], s[j])) return false;
		}
		return true;
	}

	bool isTestChar(char a)
	{
		int i = a;
		return (i>=97 && i<=122) || (i>=65 && i<=90) || (i>=48 && i<=57);
	}
	bool isEqual(char a, char b)
	{
		return a == b || int(a) == int(b)-32 || int(b) == int(a)-32;
	}
	/*
	char a = 'A';//65
	a = 'a';//97
	a = 'Z';//90
	a = 'z';//122
	a = '1';//49
	a = '0';//48
	*/

//2014-2-18 update
	bool isPalindrome(string s) 
	{
		for (int i = 0, j = s.length()-1; i < j; )
		{
			if (isalnum(s[i]) && isalnum(s[j]))
			{
				if (tolower(s[i]) != tolower(s[j])) return false;
				i++, j--;
			}
			if (i<j && !isalnum(s[i])) i++;
			if (i<j && !isalnum(s[j])) j--;
		}
		return true;
	}