【Valid Palindrome】cpp

题目：

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.

代码：

class Solution {
public:
    bool isPalindrome(string s) {
        std::transform(s.begin(), s.end(), s.begin(),::tolower);
        std::string::iterator begin = s.begin();
        std::string::iterator end = s.end();
        while ( begin < end )
        {
            if ( !::isalnum(*begin) ){
                ++begin;
            }
            else if ( !::isalnum(*end) ){
                --end;
            }
            else if ( *begin != *end ){
                return false;
            }
            else{
                ++begin;
                --end;
            }
        }
        return true;
    }
};

Tips:

1. isalnum transform函数省去了不少篇幅

2. 双指针技巧，从两头往中间逼近，不用判断奇数偶数，代码很简洁。

============================================

第二次过回文判断的题，大体思路还在，iswalnum和transform能想起来有这么个东西，具体用法记不住了。代码改了一次以后AC了。

class Solution {
public:
    bool isPalindrome(string s) {
            if (s.size()==0) return true;
            std::transform(s.begin(), s.end(), s.begin(),::tolower);
            int p1 = 0;
            int p2 = s.size()-1;
            while ( p1<p2 )
            {
                if ( !::iswalnum(s[p1]) ) { p1++; continue; }
                if ( !::iswalnum(s[p2]) ) { p2--; continue; }
                if ( s[p1++]!=s[p2--] ) return false;
            }
            if (p1>p2) return true;
            return s[p1]==s[p2];
    }
};