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125. Valid Palindrome

题目

  • 注意c/c++大小写的转换方法
  • c基本库函数isalnum(),tolower(),toupper()等基本函数的使用
 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.

解析

class Solution {
public:
  bool isDigitandApha(char ch)
	{
		if ((ch >= 'a'&&ch <= 'z') || (ch>='0'&&ch<='9'))
		{
			return true;
		}
		else
		{
			return false;
		}
	}
	bool isPalindrome(string s) {

		int size = s.size();
		int start = 0, end = size - 1;

		
		transform(s.begin(),s.end(),s.begin(),::tolower);
		string copy = s;
		while (start<=end)
		{
			if (isDigitandApha(copy[start])&&isDigitandApha(copy[end]))
			{
				if (copy[start]==copy[end])
				{
					start++;
					end--;
				}
				else
				{
					return false;
				}
			}
			else if (!isDigitandApha(copy[start]))
			{
				start++;
			}
			else if (!isDigitandApha(copy[end]))
			{
				end--;
			}
			else //都不是字符或数字
			{
				start++;
				end--;
			}
		}

		return true;
	}
};

链接：https://www.nowcoder.com/questionTerminal/b4dc0f1ee20448fca1f387fb1546f43f

bool isPalindrome(string s) {
        int i,j;
        for(i=0,j=s.length()-1;i<j;++i,--j){
            while(i<j && !isalnum(s[i])) ++i;
            while(i<j && !isalnum(s[j])) --j;
            if (i<j && tolower(s[i])!=tolower(s[j])) return false;
        }
        return true;
    }
C/C++基本语法学习 STL C++ primer