leetcode: (125) Valid Palindrome

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.

方法一：题目要求除了字母和数字外其他的符号不用考虑，定义设置前后两后两个指针，从两个方向判断；

class Solution {
public:

bool isPalindrome(string s)
{
int len =s.size();
string::iterator beg=s.begin();
string::iterator en=s.end()-1;
if (beg==en) return true;
while(beg<en)
{
if(!(isalpha(*beg)||(*beg>='0'&&*beg<='9'))) beg++;
if(!(isalpha(*en)||(*en>='0'&&*en<='9'))) en--;
if((isalpha(*beg)||(*beg>='0'&&*beg<='9'))&&(isalpha(*en)||((*en>='0'&&*en<='9'))))
{
if(*beg>='A'&&*beg<='Z') *beg+=32;
if(*en>='A'&&*en<='Z') *en+=32;
if (*beg==*en) {beg++;en--;}
else return false;
}
}
return true;
}
};

Runtime 12ms
方法二：通过库函数，较为简洁

bool isPalindrome(string s)
{
int len =s.size();
string::iterator beg=s.begin();
string::iterator en=s.end()-1;
if (beg==en) return true;
while(beg<en)
{
if(!isalnum(*beg)) beg++;
if(!isalnum(*en)) en--;
if(isalnum(*beg)&&isalnum(*en))
{
if(isupper(*beg)) *beg=tolower(*beg);
if(isupper(*en)) *en=tolower(*en);
if (*beg==*en) {beg++;en--;}
else return false;
}
}
return true;
}

Runtime 16ms