LeetCode: Valid Palindrome

最新推荐文章于 2021-05-09 05:51:44 发布

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本文介绍了一种用于判断字符串是否为回文的有效算法，该算法只考虑字母数字字符且忽略大小写。通过双指针技术从两端向中间扫描，跳过非字母数字字符，将所有字符转换为大写或小写进行比较。

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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.

class Solution {
public:
    bool isPalindrome(string s) {
        if(s.empty())
            return true;
        int i = 0, j = s.size() - 1;
        while(j > i)
        {
            while(!isNumChar(s[i]) && i < s.size())
                i++;
            while(!isNumChar(s[j]) && j > 0)
                j--;
            if(i > j)
                break;
            if(toUpper(s[i]) != toUpper(s[j]))
                return false;
            i++;
            j--;
        }
        return true;
    }
private:
    bool isNumChar(char input)
    {
        if((input >= 'a' && input <= 'z') || (input >= 'A' && input <= 'Z') || (input >= '0' && input <= '9'))
            return true;
        return false;
    }
    char toUpper(char input)
    {
        if(input >= 'a' && input <= 'z')
            return input - 0x20;
        else
            return input;
    }
};

Round2:

class Solution {
public:
    bool isPalindrome(string s) {
		int left = 0, right = s.size()-1;
		while(left < right)
		{
			if(helper(s[left]) == false)
			{
				left++;
				continue;
			}
			else if(helper(s[right]) == false)
			{
				right--;
				continue;
			}
			else
			{
				if(toLower(s[left]) == toLower(s[right]))
				{
					left++;
					right--;
				}
				else
				{
					return false;
				}
			}
		}
		return true;
	}
	bool helper(char c)
	{
		if((c >= 'A' && c <= 'Z') || (c >= 'a' && c <= 'z') || (c >= '0' && c <= '9'))
			return true;
		else
			return false;
	}
	char toLower(char c)
	{
		if(c >= 'A' && c <= 'Z')
		{
			return c + 32;
		}
		else
			return c;
	}
};

Round 3:

class Solution {
public:
    bool isPalindrome(string s) {
		if(s.empty())
			return true;
		int l = 0, r = s.size()-1;
		while(l < r)
		{
			if(!isAlphaNum(s[l]))
			{
				l++;
				continue;
			}
			if(!isAlphaNum(s[r]))
			{
				r--;
				continue;
			}
			if(toupper(s[l]) != toupper(s[r]))
			{
				return false;
			}
			l++;
			r--;
		}
		return true;
    }
private:
	bool isAlphaNum(char c)
	{
		if((c >= 'a' && c <= 'z') || (c >= 'A' && c <= 'Z') || (c >= '0' && c <= '9'))
			return true;
		else
			return false;
	}
};